categorified.netcategorified.net/LieGroups.pdf · Lie Groups Prof. Mark Haiman notes by Theo Johnson-Freyd UC-Berkeley Mathematics Department Fall Semester 2008 Contents Contents

Lie Groups

Prof. Mark Haimannotes by Theo Johnson-Freyd

UC-Berkeley Mathematics DepartmentFall Semester 2008

Contents

Contents 1

Introduction 40.1 Conventions and numbering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Lecture 1 August 27, 2008 61.1 About the course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Let’s Begin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Lecture 2 August 29, 2008 92.1 Closed Linear Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Lecture 3 September 3, 2008 12

Lecture 4 September 5, 2008 144.1 Classical compact Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.2 Classical complex semisimple Lie/algebraic groups . . . . . . . . . . . . . . . . . . . 15

Lecture 5 September 8, 2008 165.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Lecture 6 September 10, 2008 186.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Lecture 7 September 12, 2008 21

Lecture 8 September 12, 2008 248.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

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8.2 Integral Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Lecture 9 September 17, 2008 279.1 Vector fields and group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Lecture 10 September 19, 2008 3010.1 The exponential map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Lecture 11 September 22, 2008 3311.1 Universal Enveloping Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Lecture 12 September 24, 2008 3612.1 The Universal Enveloping Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Lecture 13 September 26, 2008 3813.1 U(g) is a bialgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Lecture 14 September 29, 2008 4114.1 Baker-Campbell-Hausdorff formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Lecture 15 October 1, 2008 4315.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Lecture 16 October 3, 2008 4616.1 Review of algebraic topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Lecture 17 October 6, 2008 4917.1 Two-dimensional Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4917.2 A dictionary between algebras and groups . . . . . . . . . . . . . . . . . . . . . . . . 51

Lecture 18 October 8, 2008 5218.1 More structure theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Lecture 19 October 10, 2008 5519.1 Engel’s Theorem and Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5519.2 Solvability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Lecture 20 October 13, 2008 5720.1 Solvable algebras (char = 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5820.2 Killing Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5820.3 Jordan Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Lecture 21 October 15, 2008 6121.1 Cartan’s Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6121.2 Three-dimensional Lie algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

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21.3 Casimir operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Lecture 22 October 17, 2008 6422.1 Review of Ext: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Lecture 23 October 20, 2008 6723.1 Computing Exti(K,M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Lecture 24 October 22, 2008 7024.1 Ext1

U(g)(M,N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7024.2 Ext2

U(g)(M,N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Lecture 25 October 24, 2008 7325.1 Ado’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Lecture 26 October 27, 2008 7626.1 Semisimple Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Lecture 27 October 29, 2008 7927.1 sln and sp2n over C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Lecture 28 October 31, 2008 8228.1 sl2(C) modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Lecture 29 November 3, 2008 8629.1 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Lecture 30 November 5, 2008 8830.1 Continuations of last time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8930.2 Jordan decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Lecture 31 November 7, 2008 91

Lecture 32 November 10, 2008 9432.1 Properties of root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Lecture 33 November 12, 2008 9733.1 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97


Lecture 35 November 17, 2008 10335.1 Classification of finite-type Cartan matrices . . . . . . . . . . . . . . . . . . . . . . . 103


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36.1 From Cartan matrix to Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107


Lecture 38 November 24, 2008 11138.1 Irreducible modules over g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112


Lecture 40 December 1, 2008 11440.1 A little about algebraic geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Lecture 41 December 3, 2008 11741.1 Review of Algebraic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11741.2 Algebraic Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Lecture 42 December 5, 2008 120

Lecture 43 December 8, 2008 122

Lecture 44 December 8, 2008 12444.1 Finishing up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Theo’s answers to Problem Set 1 127






Index 204

Introduction

These notes are from the class on Lie Groups, taught by Prof. Mark Haiman at UC-Berkeley in theFall of 2008. The class meets three times a week — Mondays, Wednesdays, and Fridays — from10am to 11am. MH’s website for the course is at http://math.berkeley.edu/∼mhaiman/math261A-fall08/. The course continued in the spring as Nicolai Reshetikhin’s 261B: Quantum Groups. Notesfrom that course are available at http://math.berkeley.edu/∼reshetik/math261B.html.

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http://math.berkeley.edu/~mhaiman/math261A-fall08/

http://math.berkeley.edu/~mhaiman/math261A-fall08/

http://math.berkeley.edu/~reshetik/math261B.html

I typed these notes mostly for my own benefit, although I do hope that they will be of use to otherreaders. I apologize in advance for any errors or omissions. Places where I did not understand whatwas written or think that I in fact have an error will be marked **like this**. Please e-mail me([email protected]) with corrections. For the foreseeable future, these notes are availableat http://math.berkeley.edu/∼theojf/LieGroups.pdf.

For other Lie Groups notes, you might also check out Anton Geraschenko’s notes from 2006 athttp://math.berkeley.edu/∼anton/written/LieGroups/LieGroups.pdf — those are rather compre-hensive, based on the one-semester version of this class taught then. For a very different introductionto Lie theory, see John Baez’s “Lie Theory Through Examples” seminar taught Fall 2008 (same timeas this course), with notes online available at http://math.ucr.edu/home/baez/qg-fall2008/. Evenmore recently, Peter Woit has posted an overview and application of much of this material, explain-ing “BRST Symmetry” — the primary tool for renormalizing quantum gauge theories — in a re-markably understandable way, at http://www.math.columbia.edu/∼woit/wordpress/?cat=12.

These notes are typeset using TEXShop Pro on a MacBook running OS 10.5; the TEX backend ispdfLATEX. Most of the diagrams and images are drawn with XY-pic, some live but many addedafter-the-fact. Some diagrams I didn’t get to until Winter Break, at which time I was learningpgf/TikZ, so some diagrams are in that generally superior language. The raw TEX sources areavailable at http://math.berkeley.edu/∼theojf/LieGroups.tar.gz. These notes were last updatedJanuary 23, 2009.

0.1 Conventions and numbering

Each lecture begins a new “section”, and if a lecture breaks naturally into multiple topics, I try toreflect that with subsections. At the end, I’ve included the problem sets given out as appendices,complete with my responses — be warned that these are even more likely than my notes to haveerrors, so take them with a healthy dose of salt.

Equations, theorems, lemmas, etc., are numbered by their lecture. Theorems, lemmas, and propo-sitions are counted the same, but corollaries are assumed to follow from the most recent theo-rem/lemma/proposition.

Definitions are not marked qua definitions, but italics always mean that the word is being definedby that sentence. All definitions are indexed in the index.

We use blackboard bold letters in the standard way, e.g. for the Real numbers. Indeed, we haveyet to use any particularly nonstandard notational conventions. Occasionally I will use differentconventions than MH does: I prefer ≤ for “subgroup”, and ⊆ for “subset”; 1 for the unit in a grouprather than e; K for a generic field rather than k.

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mailto:[email protected]

http://math.berkeley.edu/~theojf/LieGroups.pdf

http://math.berkeley.edu/~anton/written/LieGroups/LieGroups.pdf

http://math.ucr.edu/home/baez/qg-fall2008/

http://www.math.columbia.edu/~woit/wordpress/?cat=12

http://math.berkeley.edu/~theojf/LieGroups.tar.gz

Lecture 1 August 27, 2008

The room is very crowded. We will try to move next time.

1.1 About the course

This is the first semester of a two-semester sequences, although this course is self-contained onLie Groups and Lie Algebras. We will try to include some material on Algebraic Groups, lookingforward towards Reshetikhin’s class on Quantum Groups.

There are two textbooks: Lie Groups Beyond an Introduction by Knapp, and Linear AlgebraicGroups by Borel.

Each is missing something. The former is lacking an introduction, which we will talk about — closedsubgroups of matrix groups — and also standard background. In particular, the identification oflie groups and lie algebras is stated but not proved.

There will be no exams, grading is based on homework, but we have no grader right now. Theusual procedure is to make up a running list of problems, which will be available on the website forthe course. By the end of term there should be hundreds of them, and you should do a respectablefraction of them. This is not so definite. You can turn them in for grading, and some fraction willbe graded.

1.2 Let’s Begin

A group is a set with a multiplication satisfying axioms. A group captures the idea of symmetry,and we are generally interested in the action of groups on sets. In geometry, we’re interested inactions on spaces, which have some geometry, and you can often see a lot more of what’s going onif the symmetries themselves have a structure as a geometric space. E.g. a topological space canbe acted on by a topological group continuously. E.g. differentiable manifolds. The theory of LieGroups is the theory of groups that are differentiable manifolds.

Let’s do this in all settings, by working in diagrams.

G2 = G×G µ→ G multiplication

Gi→ G inverse

G0 = {pt} e→ G identity

We recall the notion of a cartesian product, and we should consider {pt} to be the cartesian productof zero spaces: there’s 1 way to make a map from a space to no spaces. Cartesian products arenaturally associative, etc.

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The axioms:

G×G×G1G×µ //

µ×1G��

G×Gµ

��G×G

µ // G

commutes. Other group laws are similar. E.g.

G = G× {pt} 1G×e // G×Gµ // G

is the identity map.

We will not digress into a discussion of categories, but use category theory as a language. Notevery category has products, but given two objects S and T , we call the diagram

T

S × T

<<xxxxxxxxx

""FFFFFFFFF

S

the product of S and T if the maps X → S × T are in canonical bijection to pairs of maps X → Sand X → T such that the diagram commutes:

T

X

55kkkkkkkkkkkkkkkkkk

))SSSSSSSSSSSSSSSSSS // S × T

<<xxxxxxxxx

""FFFFFFFFF

S

Officially differential geometry is a prereq for the course, although we won’t need much. But let’sassume that categories like “Topological Spaces” and “Riemannian Manifolds” exist.

A group object in a category with finite products (including the empty product, i.e. the terminalobject, which in all our geometric categories is {pt}) is an object G with maps as above, satisfyingthe group axioms written diagrammatically.

Example: A topological group is a group in Top, i.e. an abstract group whose underlying set has atopology, and such that the group operation and the inverse map are continuous. In principlewe must also require that e is continuous, but this is not a problem.

In Lie Groups, there’s a fundamental issue: do we want analytic manifolds or smooth manifolds? Itturns out it doesn’t matter: every C∞ lie group has an analytic structure. It’s easier to work in theanalytic category, although the book takes the other tact (and uses “analytic” to mean somethingelse)

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Example: A real C∞ lie group is a group object in C∞-Man, i.e. an abstract group whose underly-ing set is a smooth manifold, and the maps are smooth maps. We say that a map φ : X → Yis smooth if for any f : Y → R smooth, the map f ◦ φ : X → R is smooth.

Example: A real (analytic) lie group is a group object in Man/R. We will take analytic manifoldsto be the basic notion.

Question from the audience: Requiring manifolds to be analytic messes up partitions of unity?Answer: Yes, but this doesn’t mess us up too much. Partitions of unity allow us to reduce globalquestions to local questions. But we won’t need them, because we will only ask local questionsanyway.

Example: A complex lie group is a group object in Man/C. Manifolds over C are hard, becauseCn is very rigid. A differentiable function on C is already analytic, so we need to say all sortsof things like “maps from between open set in the manifold to open sets in Cn”. There is anotion of “almost complex manifold”, but any such group is actually an analytic lie group.

Example: A (linear) algebraic group is a group in the category of (affine) algebraic varieties overk = k. We will always take our ground field to be k = C. We will talk more carefully aboutthis later. For our purposes, “algebraic variety” means “reduced algebraic variety over analgebraically closed field”. Question from the audience: But not irreducible? Answer:No, we never demand connectedness. And we really mean the underlying set of closed points:for our purposes, an algebraic variety really is a solution set of a polynomial. So we haveX ⊆ kn, the solution set to some ideal I(X) ⊆ k[x1, . . . , xn]. Then the coordinate ring on Xis k[x1, . . . , xn]/I(X) and a map X → Y is a map k[x1, . . . , xn]/I(X) ← k[y1, . . . , ym]/I(Y ).So we can reverse all our arrows, and work in the category of finitely-generated algebras overk. In any case, this gives the notion of “Hopf algebra”.

Algebraic varieties can have singular points, but it’s a basic theorem that not every point is singular.But if it’s a group, then it acts transitively on itself, so an algebraic group cannot have singularpoints. Thus, we’ve listed a hierarchy of notions.

Incidentally, what does it mean for G to act on X? A group action is a map G×X ρ→ X such thatsome diagrams commute.

G×G×Xµ×1X

''NNNNNNNNNNN

1G×ρwwppppppppppp

G×Xρ

''NNNNNNNNNNNN G×X

ρwwpppppppppppp

X

Example: GLn(C). But to be an algebraic group, the maps must all by polynomials. And theinverse is not obviously a polynomial. Well, X−1 = adjX/ det(X). This acts algebraically onCn. It also acts on projective space P(Cn) = Pn−1(C).

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Lecture 2 August 29, 2008

Effective next week, we will be in 145 McCone, says the online schedule of classes.

Today we look at a number of examples, but begin with some theory. The most straightforwardway to get examples are as closed subgroups of matrix groups, either over C or R. Some subgroupsare formed using polynomials — if over C, these give algebraic groups — but other subgroups arenot algebraic.

2.1 Closed Linear Groups

A closed linear group is a subgroup of GLn (over C or R). We demand it be closed as a topologicalsubspace. But it’s not even obvious that a closed subgroup is Lie, i.e. that it’s a submanifold.

The critical notion is the matrix exponential: we have a map from Mn → GLn by A 7→ eA. Thereare a number of equivalent definitions, but here’s one:

eAdef=∑n≥0

An

n!(2.1)

We can also define it aseA

def= etA∣∣∣t=1

(2.2)

where the above solves the differential equation

d

dtetA = AetA (2.3)

Another definition:eA

def= limn→∞

Å1 +

A

n

ãn(2.4)

The exponential satisfies similar relations as in commutative land. E.g. if AB − BA = 0, theneAeB = eA+B. Note: we’ve introduced the bracket [A,B] def= AB −BA.

When A and B don’t commute, we can add counterterms:

etA = 1 + tA+12t2A2 +O(t3) (2.5)

etB = 1 + tB +12t2B2 +O(t3) (2.6)

etAetB = 1 + tA+ tB +12t2A2 +

12t2B2 + t2AB +O(t3) (2.7)

et(A+B) = 1 + tA+ tB +12t2A2 +

12t2B2 + t2

AB +BA

2+O(t3) (2.8)

etAetB = et(A+B)+ 12t2[A,B] +O(t3) (2.9)

= et(A+B)e12t2[A,B] +O(t3) (2.10)

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So what’s happening? The group law in GLn sees the addition law of matrices:

A+B =d

dt

∣∣∣∣t=0

etAetB (2.11)

Moreover, the group commutator:

etAetBe−tAe−tB = et2

2[A,B]et(A+B)e−t(A+B)e

t2

2[A,B] +O(t3) (2.12)

= et2[A,B] +O(t3) (2.13)

So the corollary:

[A,B] =12d2

dt2

∣∣∣∣∣t=0

etAetBe−tAe−tB (2.14)

We can take logs, too. But this is, of course, hairy: even in C, we need branch cuts, etc. But wecan restrict to a small ball around 1. Question from the audience: What you’re saying is thatexp is onto? Answer: Well, it is onto, but that’s because every matrix can be written in canonicalform. But we take a particular log near 1:

logA def= −∑n>0

(1−A)n

n(2.15)

which converges for |1 − eigenvalues| < 1, and it’s an inverse of exp in the neighborhood. Solocally this is an analytic isomorphism between a neighborhood of 0 ∈ Mn and a neighborhood of1 ∈ GLn.

Furthermore, it’s a corollary of the formula that the product of exponentials is almost the expo-nential of the sum that if we break Mn = V ⊕W , then we can map (v, w) 7→ exp(v) exp(w), whichis not the same as exp(v + w), but it’s close. This also defines an analytic homeomorphism ofneighborhoods of 0 and 1 as above. You can jazz this up more, even taking a whole coordinatesystem, and get lots of different analytic isomorphisms.

So, suppose we have a closed subgroup of GLn. We want to take its log, and see what kind ofsubgroup of Mn we get. But that’s not exactly what we want, e.g. we might have taking a discretesubgroup. Because even if H < GLn is the trivial subgroup, then the matrices that exponentiateto 1 are all things with eigenvalues 2πi. So we really just want to study the parts of the subgroupnear the origin.

Question from the audience: Did you say exp is onto? Even over R? Answer: No, it’s ontoover C, but not R, and not in general Lie groups. But it doesn’t matter.

What are the closed subgroups of (R,+)? Well, there’s R, 0, and Zλ. Question from theaudience: What about Q? Answer: Not closed. And we want to think of Z and 0 as basicallythe same.

So, given a closed H < GLn, define the Lie Algebra of H to be the set Lie(H) = {X ∈Mn : eRX ⊆H}. Question from the audience: It’s enough to take a small neighborhood? Answer: Ofcourse, but it will be a subgroup of R, and there aren’t many of those.

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Question from the audience: Why does it have the same dimension as H? Answer: I haven’tsaid it does yet. It’s not even obvious that it’s a submanifold of Mn. We will show that in a nbhdof 0, the exponential map takes Lie-H exactly onto H. Then it must be a submanifold, because ofcourse anything that looks like a line in some coordinate must be a submanifold. But we havne’teven shown that it’s a subspace.

Proposition 2.1: (a) Lie(H) is a R-subspace. (b) Lie(H) is closed under [, ] (i.e. it’s a Liesubalgebra of Mn).

Example: U1 < GL1(C). Then Lie(U1) = iR. This is not a complex algebraic subgroup, becausewe needed to use complex conjugation to define U1 = {z : zz = 1}.

Proof of Example:

Ok, so if X,Y ∈ Lie(H), then etXetY ∈ H, and if t is small, this is close to et(X+Y ). So whatwe do is look at

ÄeX/neY/n

än ∈ H. But as n → ∞, this isÄ1 + X+Y

n +O(1/n2)än → eX+Y .

So do this decorated by ts, and we see that X + Y ∈ H. This proves (a).

For (b), we takeÄetX/netY/ne−tX/ne−tY/n

än2

−→n→∞

Ç1 +

t2

n2[X,Y ] +O(1/n3)

ån2

→ et2[X,Y ]

Therefore [X,Y ] ∈ Lie(H). �

Example: [

Important Counterexample]Take the torus T = U1×U1 ∈ GL2(C). I.e. T =®ñ

xy

ô: |x| = |y| = 1

´.

So this is some rectangle, and the Lie(T ) =®ñ

ixiy

ô: x, y ∈ R

´. Ok, so let’s pick a λ, and

look at the Lie algebra®ñ

itiλt

ô: t ∈ R

´. Exponentiating this, if λ = r/s ∈ Q, gives a nice

closed subgroup of T , namely when xr = ys. But if λ 6∈ Q, then the exponential of this Lie algebrais dense in T . This is a subgroup, but not a closed subgroup of T . The point is: being a Liesubalgebra is not enough to guarantee naively that the exponential is a Lie subgroup. What weactually have is a continuous group homomorphism (R,+)→ T , which is locally on the domain asmooth embedding.

So what you should mean by a subgroup in the category of manifolds is another group with acontinuous smooth embedding. So with a more sensible definition of “Lie subgroup”, we do havethat Lie subalgebras correspond to Lie subgroups. So the conclusion is that a Lie subgroup neednot have a closed image.

Question from the audience: How systematic can we be talking about non-closed things?Answer: It’s worse than I made it out to be. Take T as a group, with the discrete topology.Then the identity map discrete → smooth is a smooth map of manifolds. Question from theaudience: But that’s not second-countable. Answer: Sure, so one solution is to require that

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manifolds be second-countable, which is designed to eliminate this pathology. But in a category,the right way to say “subgroup” is as a map, and so in this class, T with the discrete topology isa perfectly good disconnected Lie subgroup. And this is no problem for the theorem: Lie algebrasonly know about connected Lie groups.

Lecture 3 September 3, 2008

We continue the discussion of closed subgroups of GLn, and maps thereof.

Recall from last time:

Lemma 3.1: The exponential map is a homeomorphism in a neighborhood of the identity, andmoreover breaking up the Lie Algebra into direct sum this still works. Saying this again: ifMn = V ⊕W as a Real vector space, then there exists a(n open) nbhd U of 0 ∈ Mn and anbhd U ′ of 1 in GLn s.t. (v, w) 7→ exp(v) exp(w) is a homeo U → U ′.

This is because exp(v) = 1+v+small, so to first approximation the above is the identity map.

Let’s take a closed subgroup H < GLn, which can be real, even if GL is complex. Then defineLie(H) def= {x : expRx ⊆ H}.

Question from the audience: We had the discussion of the torus and the dense real line.Shouldn’t that mean we change the definitions of closed subgroups? Answer: Well, it’s a factthat not every closed Lie subalgebra gives a closed Lie subgroup. But we will have a statement like“Every Lie algebra homomorphism lifts to a Lie group homomorphism”, although there will stillbe conditions (simply connected, etc.) Question from the audience: So how do you define theLie algebra of that subgroup? Answer: Well, taking the topology of R as embedded irrationallyin the torus gives the wrong topology: it’s not a manifold with the induced topology. The pointis the notion of submanifold should not be the most naive definition: it’s not a subspace with theinduced topology.

In any case, we want to prove something about closed subgroups, which will rule out this picture(of R ↪→ T 2 irrationally.)

Proposition 3.2: There exists a nbhd U 3 0 ∈ Mn and U ′ 3 1 ∈ GLn s.t. exp : U ∼→ U ′, and s.t.Lie(H) ∩ U ∼→ H ∩ U ′. **We write f : A ∼→ B for the statement “f : A → B is aniso”.**

Proof of Proposition 3.2:

Fix a complement W ⊆Mn s.t. Mn = Lie(H)⊕W .

Oh, we’re missing a page of notes. First:

Lemma: Given any linear subspace W ⊆ Mn, is 0 is a limit pt of {w ∈ W : exp(w) ∈ H},then W ∩ Lie(H) 6= 0.

12

Proof of Lemma:

Fix a Euclidian norm on W . Say w1, w2, · · · → 0. Then wi/|wi| are on the unit sphere,which is compact, so passing to a subsequence, we can assume that wi/|wi| → x wherex is a unit vector. Well, the norms |wi| are tending to 0, so wi/|wi| is a large multipleof wi. Let’s approximate this: let ni = d1/|wi|e, and then niwi ≈ wi/|wi|. So niwi → x.And expwi ∈ H, so exp(niwi) ∈ H, and H is a closed subgroup, so expx ∈ H.

We could have taken any unit ball. By the same argument, we could have taken anyball of radius r to conclude that exp(rx) is in H, hence x ∈ Lie(H). �

In any case, returning to the proof of the proposition, we can find W ⊇ V 3 0 s.t. {v ∈ V :exp(v) ∈ H} = 0. Then we map (x,w) 7→ exp(x) exp(w). Then exp(x) ∈ H, and by choosinga suitably small nbhd, we can assure that expw 6∈ H unless w = 0.

And (x,w) 7→ exp(x) exp(w) is almost the map (x,w) 7→ exp(x + w). There’s a change ofcoordinates that fixes this, completing the proof. �

This is telling us that the Lie algebra is locally a picture of the group. We haven’t defined a manifoldor a submanifold, but this proposition tells us that we definitely have a submanifold. I.e.:

Corollary 3.2.1: H is a submanifold of GLn of dimension = dim Lie(H).

Corollary 3.2.2: exp(Lie(H)) generates the identity component H0 of H.

Example: Orthogonal matrices can have determinant ±1, so at least two connected components.Exercise: the orthogonal matrices of determinant +1 are connected.

In general, any closed subgroup will have a distinguished subgroup, the identity component, andthe other components are its cosets. The identity component is normal, because to get to anyother component, we just pick an element, and multiply on the left or right by it. So the pictureof subgroups H of GLn is that each has an identity component H0 and a discrete group that’sH/H0, so H is some extension of a connected thing by a discrete thing. And we said that exp LieHgenerates H0.

By the way, any open subgroup is also closed. Its cosets are also open, and it’s the complement ofthe union of its cosets. This is true in any topological group.

Let γ : R→ GLn be a smooth curve, i.e. the matrix entries in γ(t) are smooth functions, and let’sassuming that γ(0) = 1 ∈ GLn. **I’m going to use 1 to mean the identity of any group.On the board MH writes e, or In for the identity in GLn.** Then γ′(0) is the tangent vectorto the curve at the origin. Well, the tangent space is T1GLn = T0Mn = Mn. And we define thetangent space T1H

def= {γ′(0) s.t. γ : R→ H, γ(0) = 1} ⊆Mn.

Corollary 3.2.3: Lie(H) = T1H.

We will use this as a definition of the Lie algebra of an abstract Lie group.

Some examples: The classical Lie groups come in a few flavors. For us, there are two flavors, but

13

if you are more serious about harmonic analysis, you might think there are more flavors. Thereare hugely many Lie groups, and we will talk about the semisimple Lie groups. The flavors wecare about are the compact ones and the complex ones. These are classified by the same data:any complex semisimple Lie group contains a real compact one. There are also real non-compactgroups, e.g. SLn(R). These are interesting, e.g. the book by Lang on SL2. The point is that therereally is a third flavor: the other ones.

Example: GLn(C), GLn(R). The former isn’t quite semisimple, but the special linear groupSLn(C) def= {A ∈ GLn(C) s.t. detA = 1} is. It has a compact cousin. The unitary groupUn

def= {A ∈ GLn(C) s.t. AA∗ = 1}. Recall that A∗ is the conjugate transpose of A. Theseneed not have determinant 1, by multiplying by eiθ. And we have the special unitary groupSUn

def= {A ∈ Un s.t. detA = 1}. Indeed, LieGLn(C) = Mn(C). And Lie(Un) = {X ∈Mn s.t. X = −X∗}, i.e. the skew-Hermetian matrices. It’s not entirely obvious that thisis closed under Lie bracket. Imposing the S condition: the eigenvalues of exp are exp ofthe eigenvalues, so LieSLn(C) = {X ∈ Mn(C) s.t. trX = 0}, and Lie(SUn) are the skew-Hermetian matrices of trace 0.


Last time, we talked about two important flavors of Lie groups: (semi-simple) compact Lie groupsand (semi-simple) complex Lie groups. In the compact case, almost all are semi-simple: a compactLie group with no center is semi-simple. The complex ones are more complicated. We will talkabout the classical constructions, and see that there are a few families and some sporadic ones.

4.1 Classical compact Lie groups

Of the classical family, the first is naturally a family of compact subgroups of GLn(R), the secondof GLn(C), and the third of GLn(H). Of course, we can consider C ∼= R2 and H ∼= C2, althoughthe latter is hairy, because it’s a different space from the right or the left.

Recall the quaternions H = {x+ iy+jw+kz} where the multiplication is i2 = j2 = k2 = ijk = −1.This is a non-commutative division ring. We can define the complex conjugate in the obviousway: i = −i, etc. This is an anti-automorphism: αβ = βα. We have a Euclidean norm given by‖ζ‖2 = ζζ.

So, let’s say that x ∈ Rn,Cn,Hn is a column vector. We can sum the square norms of the vectorentries: ‖x‖2 = xTx. More generally, for any matrix X, define its Hermetian conjugate to beX∗ = XT . Then (XY )∗ = Y ∗X∗. We have a subgroup of matrices s.t. X∗X = 1, since (XY )∗XY =Y ∗X∗XY = Y ∗Y = 1 in the subgroup.

We define the special orthogonal group: SO(n) def= {X ∈ Mn(R) : X∗X = 1 and detX = 1}. Therestriction on the determinant isn’t much: if X∗X = 1, then detX = ±1, so we are picking out

14

just the connected component, without changing the Lie algebra. SU(n) def= {X ∈Mn(C) : X∗X =1 and detX = 1}. Here the determinant condition gets rid of the central elements. Well, thecenter is discrete. Sp(n) def= {X ∈ Mn(H) : X∗X = 1}. There’s no determinant condition, fortwo reasons: there’s no good determinant over H, and the center is already finite and the group isconnected.

The condition that X∗X = 1 says that all columns x have ‖x‖2 = 1, and that the columns arelinearly independent. So in particular the columns lie on the unit ball. Since these are solutions ofa continuous function, the groups are closed subsets of the (compact) unit ball, so all these groupsare compact.

In any case, this list is almost all the compact subgroups. Well, there’s quotients of the groupsby their centers, there’s extensions by the circle, and things like the spin group (double cover ofSO(n)). Oh, and there’s finitely many others.

We need to fix our definition, since last time we didn’t consider H. Well, Mn(H) y Hn on theleft, which commutes with the H action from the right. **I’d say Mn(H)HnH.** Let’s pick a copyof C inside H: C = R ⊕ iR ⊆ H. Then we can think of H = {w + jz : w, z ∈ C} with theconditions that j2 = 1 and jz = zj. So H ∼= C2 as a (right) C-vector space, with the basis {1, j}.So Hn ∼= C2n = Cn ⊕ jCn as right C-v-space.

Now, let’s say A = W + jZ ∈Mn(H) where W,Z ∈Mn(C). A acts on Hn ∼= C2n by block 2n× 2nmatrix:

A =ñW −ZZ W

ôSo in particular

Jdef= j1Hn =

ñ0 −1Cn

1Cn 0

ôIn any case, if X ∈M2n(C) ⊇Mn(H), then X ∈Mn(H) if and only if JX = XJ .

Remark: This embedding of H as C-matrices is the same as the way we can think of C as R-matrices.

4.2 Classical complex semisimple Lie/algebraic groups

We can define these over any field, but it’s best to work over an algebraically closed field.

SL(n,C) def= {X ∈ GLn(C) s.t. detX = 1} (4.1)

SO(n,C) def= {X ∈ SLn(C) s.t. XTX = 1} (4.2)

Sp(n,C) def= {X ∈ GL2n(C) s.t. XTJX = J} (4.3)

We have no conjugations, because we want these to be algebraic groups.

15

The last is probably new, and not obviously a group. But take any bilinear form 〈x, y〉 = xTAy

for some matrix A. Then if g ∈ GLn(C), then g〈x, y〉 def= 〈g−1x, g−1y〉 = xT (g−1)TAg−1y, so thestabilizer of 〈, 〉 is all g such that gTAg = A. And SO(n,C) is the stabilizer of the nondegeneratesymmetric form (only one over C up to conjugation), and Sp(n) fixes a nondegenerate antisymmetricform, which only exists on even dimensional vector spaces.

We compute Lie algebras:

GLn(H) � � open // Mn(H) � � R-linear subspace // M2n(C)

soLie(GLn(H)) = Mn(H) = {X ∈M2n(C) s.t. JX = XJ} (4.4)

Moreover, det eX = etrX , so

Lie(SL(n,C)) = {X s.t. trX = 0} def= sl(n,C) (4.5)

Well, (eX)T = e(XT ), so

Lie(SO(n,C)) = {X s.t. X +XT = 0} def= so(n,C) (4.6)

**This line filled in later from Hwajong Yoo:** Moreover, Sp(n,C) = {A : AT = JA−1J−1},and eJ−XJ

−1= Je−XJ−1, thus XTJ = −JX iff XT = −JXJ−1 iff (eX)T = eX

T= eJ(−X)J−1

=Je−XJ−1 = J(eX)−1J−1. Thus:

Lie(Sp(n,C)) = {X s.t. XTJ = −JX} (4.7)

How about the real ones? We look at the fixed-points of the above under complex conjugation.

Lie(SO(n,R)) = {X ∈Mn(R) s.t. XT +X = 0} (4.8)Lie(U(n)) = {X ∈Mn(C) s.t. X∗ +X = 0} (4.9)

Lie(SU(n)) = {X ∈ Lie(U(n)) s.t. trX = 0} (4.10)

Notice that dim LieU(n) = n2, dim LieSU(n) = n2 − 1.


Recall from last time: We write X for the complex conjugate of a matrix X, and XT for thetranspose; then X∗

def= XT . We let J ∈M2n(C) be the block matrix

J =ñ

0 −11 0

ô∈M2(Mn(C)) = M2n(C)

16

Then we have the following “classical” closed linear groups:

Group Group Algebra AlgebraName Description Name Description dimR

Com

pact SO(n,R) {X ∈Mn(R) s.t. X∗X = 1, detX = 1} so(n,R) {X ∈Mn(R) s.t. X∗ +X = 0}

(n2

)SU(n) {X ∈Mn(C) s.t. X∗X = 1,detX = 1} su(n) {X ∈Mn(C) s.t. X∗ +X = 0, trX = 0} n2 − 1Sp(n) {X ∈Mn(H) s.t. X∗X = 1} sp(n) {X ∈Mn(H) s.t. X∗ +X = 0} 2n2 + n

GLn(H) {X ∈ GL2n(C) s.t. JX = XJ} gln(H) {X ∈M2n(C) s.t. JX = XJ} 4n2

=®ñ

W −ZZ W

ô´=®ñ

W −ZZ W

ôs.t. W,Z ∈Mn(C)

´

Com

plex SL(n,C) {X ∈Mn(C) s.t. detX = 1} sl(n,C) {X ∈Mn(C) s.t. trX = 0} 2(n2 − 1)

SO(n,C) {X ∈Mn(C) s.t. XTX = 1, detX = 1} so(n,C) {X ∈Mn(C) s.t. XT +X = 0} n(n− 1)Sp(n,C) {X ∈Mn(C) s.t. XTJX = J} sp(n,C) {X ∈Mn(C) s.t. XTJ + JX = 0} 2

(2n+12

)**I couldn’t quite keep up with the table live. This is constructed later from my jottednotes; some is still missing.**

Proposition 5.1: Via Mn(H) ↪→M2n(C), we have

Sp(n) = GLn(H) ∩ U(2n)= GLn(H) ∩ Sp(n,C)= U(2n) ∩ Sp(n,C)

And, of course, the corresponding statements for Lie algebras.


Nothing deep. You check the above exact descriptions all imply each other. Uses thatMn(H) ↪→M2n(C) is a ∗-embedding. The end is an exercise: Sp(n) is a compact real form ofSp(n,C), in that the latter is the direct sum of the former and i times the former. �

By varying the above statement, you get similar observations.

5.1 Homomorphisms

The inclusions above are, of course, group homomorphisms, and give homomorphisms of the corre-sponding Lie algebras. More generally, homomorphisms of groups give homomorphisms of algebras,and conversely in the connected case the Lie algebra homomorphism determined the group homo-morphism.

Recall: if H < GLn is closed, then Lie(H) = {X ∈ Mn s.t. eRX ⊆ H} = T1(H) = {γ′(0) s.t. γ :R → H smooth , γ(0) = 1}. Then H y H by gh = ghg−1, fixing 1. So H y T1H. If γ(t) =

17

1 + tY + O(t2), where Y = γ′(0) ∈ T1H, then gγ(t)g−1 = 1 + tgY g−1 = O(t2), so the action isg · Y def= Ad(g)Y def= gY g−1.

Calculating further, let’s act with a curve, and calculate Ad(γ(t))Y . We know γ(t) = 1+tX+O(t2),so (γ(t))−1 = 1− tX +O(t2), so

d

dt

∣∣∣∣t=0

Ad(γ(t))Y =[γ′(0), Y

](5.1)

Let φ : H → G be a smooth homomorphism of closed linear groups. Then φ(1) = 1, so dφ : T1H →T1G by X 7→ (φ(1 + tX))′(0). The diagram of actions commutes:

H

φ

��

y T1H

dφ��

G y T1G

This is to saydφ(Ad(h)Y ) = Ad(φ(h)) dφ(Y ) (5.2)

by an easy calculation. Thus dφ [X,Y ] = [dφX, dφY ], so dφ is a Lie algebra homomorphism.

No, we also get along with the exponential map.

Lie(H)exp //

dφ

��

H

φ

��Lie(G)

exp // G

So if H is connected, then dφ determined φ. Then dφ is an isomorphism if and only if φ is alocal isomorphism. It’s not necessarily an isomorphism, and indeed not every algebra homo liftsto a group homo (it does in the simply-connected case), e.g. R versus U(1). Nevertheless, algebrahomomorphisms are a good way to detect group homomorphisms.

Example: Let’s compare Lie(SU(2) = {X s.t. X = −X∗, trX = 0} =®ñ

ia b+ ic−b+ ic −ia

ôs.t. a, b, c ∈ R

´with Lie(SO(3)) =

0 a b−a 0 c−b −c 0

s.t. a, b, c ∈ R

. So each is three-dimensional, and in

fact has a basis {X,Y, Z} where the bracket of any two is ± the third. It turns out there isa map SU(2)→ SO(3), but constructing it is hard.


We’ve finally posted a handful of problems on the website for you to enjoy or not enjoy, in orderto get yourself a feeling for some examples. Today we start with general theory. In particular,

18

we need to talk about manifolds, which come in multiple flavors: real or complex, analytic orC∞ (which isn’t a difference over C, but there is something called an “almost complex manifold”,which is like a C∞ but not necessarily analytic complex manifold). You could take any of thesefor a definition, and the book makes particular choices. But it turns out that any C∞ Lie groupis analytic, and any almost complex Lie group is complex. So sometimes you make one choiceor another for ease: we will talk about real C∞ manifolds today, but the reader can make theappropriate substitution.

6.1 Manifolds

The idea is a topological space, with enough structure that you can tell what functions are smooth.So you put on coordinate charts which know such things.

So, take a (Hausdorff) topological space X. A chart is an open U ⊆ X and a homeomorphismφ : U → V ⊆

openRn. Well, Rn has coordinates xi, and ξi

def= xi ◦ φ are local coordinates on the

chart.

Now we need a notion of when two charts are compatible, so that they give the same notions ofsmoothness. Charts (U, φ) and (U ′, φ′) are compatible is on U ∩ U ′ the ξ′i are smooth functions ofthe ξi and conversely.

U

φ

��

U ∩ U ′φ

yyssssssssssφ′

&&LLLLLLLLLL U ′

φ′

��V V ⊇W

φ′◦φ−1// W ′ ⊆ V ′ V ′

φ′ ◦ φ−1 : W →W ′ is smooth with smooth inverse (6.1)

An atlas A on X is a covering by pairwise compatible charts.

We don’t want the atlas to be part of the definition.

Lemma 6.1: If U and U ′ are compatible with all charts of A, then they are compatible with eachother.

Corollary 6.1.1: Every atlas has a unique maximal extension.

So we define a manifold to be a Hausdorff topological space with a maximal atlas.

Again, we can vary the words “real” and “smooth” to “complex” or “analytic” to get other no-tions.

Question from the audience: Do you need Zorn’s lemma to construct a maximal atlas? An-swer: No. We don’t need to make choices when deciding which charts go in, because once we havean atlas, its maximal atlas is unique.

19

We don’t want to think about maximal atlases, because they’re ugly, and instead we tend to justgive an atlas, and then maximize it if needed. But the above definition still isn’t great, because itpredates the notion of a “sheaf”.

A function f : U → R (where U is open in X) is smooth if it is smooth on local coordinates in allcharts. So we let S(U) = {smooth fns on U}. Then we have the sheaf axioms:

1. if V ⊆ U and f ∈ S(U), then f |V ∈ S(V ), and

2. if U =⋃α Uα and f : U → R such that f |Uα ∈ S(Uα) for each α, then f ∈ S(U).

Technically, these are only the axioms for a “sheaf of functions”.

Now a better way to define a manifold is as a Hausdorff space X with a sheaf of functions S, s.t.there exists a covering of X by open sets U such that (U,S|U ) is isomorphic as a space with a sheafof functions to (V,SRn |V ) for some V ⊆ Rn open.

This definition will work better for, e.g., algebraic groups. But still we will work in local coordinatesa lot, so that local statements become statements about Rn.

What’s screwy about non-Hausdorff spaces? You’d think that, since it’s locally Rn, that’s enough,but Hausdorffness is a global property. For example, the real line with a double point should notcount as a manifold. It has a bug: a smooth function defined on both points must be equal. Andwe can’t integrate vector fields to smooth curves near the double point, because it doesn’t knowwhich point to go through.

If X and Y are smooth manifolds, then a smooth map f : X → Y is a continuous map such that forall U ⊆ Y and g ∈ S(U), then g ◦ f ∈ S(f−1(U)). It’s more or less obvious that smooth maps canbe composed. Moreover we have a product: X × Y is a manifold with charts U × V . Dimensionsadd when manifolds are multiplied. This definition gives the categorical product, by just checkingthe corresponding statement about Rn: if a map to Rn+m has its first n and last m coordinates allsmooth, then all coordinates are smooth.

So we have a category Man, and a Lie group is a group object in Man. Let’s be more precise: a realLie group or complex Lie group is a group object in the analytic category of real or complex mani-folds. Of course, we have embeddings of categories: Man/C ↪→ Mananalytic/R ↪→ ManC∞/R.

How do we define a Lie algebra? What’s the tangent vector to a curve? Well, let γ : U → Rn besmooth with 0 ∈ U ⊆ Rn and γ(0) = x ∈ Rn. Then γ(t) = (ξ1(t), . . . , ξn(t), and we can define thetangent vector to be γ′(0) = (ξ′1(0), . . . , ξ′n(0)). But we want to be less coordinatefull. Well, thechain rule gives us, for all f smooth:

(f ◦ γ)′(0) =∑ ∂f

∂xi

∣∣∣∣x· ξ′i(0) (6.2)

and these define the derivatives of γ, by taking f to be coordinate functions.

So, on a manifold M with p ∈ M and γ1, γ2 : R → M s.t. γ1(0) = γ2(0) = p, then we say that γ1

and γ2 are tangent at p if (f ◦ γ1)′(0) = (f ◦ γ2)′(0) for all smooth f on a nbhd of p. (I.e. for all

20

f ∈ limU3p S(U) def= Sp.) Each equivalence classes is called a tangent vector.

Another way to do it is to consider the tangent vector to be the linear functional Dγ : f 7→ (f◦γ)′(0),which is a linear map Sp → R. But this satisfies a Liebniz rule:

Dγ(fg) = Dγf g(p) + f(p)Dγg (6.3)

Thus it is a point derivation at p.

Proposition 6.2: Every point derivation D at p comes from some smooth curve, i.e. it is Dγ forsome γ going through p.


This is not an abstract statement, but something local, so it’s about Rn. So we just need tocheck that this is true about derivations on Rn at 0.

We claim that any smooth function f in a nbhd of 0 ∈ Rn can be written as f(x) = f(0) +∑xigi(x). I.e. every smooth function that vanishes at the origin is in the ideal generated by

the xi. You can prove this e.g. by induction. Then of course gi(0) are the partials of f , so weget a curve which gives us the derivation D, take γ(t) = (D(x1), . . . , D(xn))t, and the chainrule checks out. �


Last time we defined three variants of “manifold”: smooth, analytic, holomorphic. So that we cansave time, we write “s/a/h” for “smooth, analytic, or holomorphic, respectively”. Last time wedefined a s/a/h map f : M → N . Moreover, we said that a Lie group over R/C is a group ina/h−Man. We defined TpM in general.

We would like some examples of Lie groups. So far we have examples of closed subgroups of GLn.We will discuss these more, but first we want the correct notion of “subobject”, which will end upmeaning “locally immersed submanifold”.

So, let M be a manifold, and N a topological subspace with the induced topology. Locally Mlooks like Rm, and we want a condition on N that it looks like Rn ↪→ Rm locally. So, suppose that∀p ∈ N , there’s a chart U 3 p in M with coordinates {ξi}mi=1 : U → Rm such that U ∩ N = {q ∈U s.t. ξn+1(q) = · · · = ξm(q) = 0}. Then it’s clear that U ∩ N is a chart on N with coordinatesξ1, . . . , ξn.

Proposition 7.1: • N is a manifold with an atlas given by {U ∩N}.

• The sheaf of smooth function SN is the sheaf of functions that are locally (on N) re-strictions of smooth functions on M .

• N ↪→M is s/a/h.

21

• Any s/a/h f : Z →M with f(Z) ⊆ N defines a s/a/h f : Z → N .

The last two conditions are a universal property characterizing N .


Completely obvious in local coordinates. �

Question from the audience: What happens if you have an embedding I wouldn’t think of assmooth. E.g. a cusp. Answer: We will talk about more examples. But you cannot find a chartaround this cusp as above.

We say that N ↪→ M is an immersed submanifold if it satisfies the conditions of 7.1. We say thata map Z →M is an immersion if it factors as Z ∼→ N ↪→M with N ↪→M an immersion.

Proposition 7.2: If N ↪→M is an immersed submanifold, then N is locally closed.

Proposition 7.3: Any closed linear group H ≤ GLn is an immersed analytic submanifold, and ifLie(H) is a C-subspace of Mn(C), then H is a holomorphic submanifold.


Look at the following diagram:

0� _ 1� _Mn ⊇ U

exp // V ⊆ GLnlog

oo

Lie(H) ∩ U?�

OO

//H ∩ V

?�

OO

oo

(7.1)

This defines a chart of the identity in H, and this chart can be moved anywhere else we needit. �

Remark: This picture identifies Lie(H) ∼= T1H.

But we would like more subobjects.

Example: f : R → R2 by t 7→ (t2, t3). This is smooth, injective, and a homeo onto its image(topologically it’s an immersion), but it’s not an immersed submanifold. This is not entirelyobvious. Question from the audience: Why is that smooth? Answer: Because t2 and t3

are smooth maps. What’s wrong with it is its derivative is bad: it comes to a halt. While it’shalted, it can dance around and come out in a different way. What’s bad is that g = y2 − x3

isn’t part of a coordinate system, because ∂g/∂x and ∂g/∂y both vanish at 0. To get animmersed smooth in R, you should take an equation that is. (We haven’t showed that thereis no good function that defined this curve.)

Example: Here’s another map that’s not an immersion: R → R by t 7→ t3; its inverse is t 7→ 3√t,

which is not smooth.

22

Example: R r {pt} → R2 as a curve that would cross itself if it weren’t missing the point. Thisis not an immersion, but it is a local immersion. It is not homeomorphic to its image.

Question from the audience: Is our notion of “immersion” what I’ve seen “embedding” usedas? Answer: Yes. There’s no consensus on the meaning of these words. In this talk, we’ll use“immersion” as previously defined, and “local immersion” as what we want.

Example: R → T 2 with irrational slope. This is a local immersion and also a group homomor-phism. We want to accept this as a legitimate Lie subgroup.

We now return to our mini-course in differential geometry. Let f : M → N be s/a/h mappingp 7→ q. We want to define the differential of f at p to be a map (df)p : TpM → TqN . There areseveral definitions, which are easily seen equivalent in local coordinates:

1. If [γ] ∈ TpM is represented by the curve γ, then (df)p(X) def= [f ◦ γ].

2. If X ∈ TpM is a point-derivation on SM,p, then (df)p(X) : SN,q → R (or C) is defined byψ 7→ X[ψ ◦ f ].

3. In coordinates, p ∈ U ⊆open

Rm and q ∈W ⊆open

Rn, then locally f is given by f1, . . . , fn smooth

functions of x1, . . . , xm. The tangent spaces to Rn are in canonical bijection with Rn, and alinear map Rm → Rn should be presented as a matrix:

Jacobian(f, x) = J(f, x) def=∂fi∂fj

(7.2)

The best way to say the differential is to pull the tangent bundle of N back to M along f , andthen think of df as a smooth map of tangent bundles.

We have the chain rule: if Mf→ N

g→ K, then d(g ◦ f)p = (dg)f(p) ◦ (df)p.

Theorem 7.4: Inverse Mapping Theorem

Clasically: Given s/a/h f1, . . . , fn : U → R where p ∈ U ⊆open

Rn, then f : U → Rn maps

some nbhd V 3 p bijectively to W ⊆open

Rn with s/a/h inverse iff det J(f, x) 6= 0.

Coordinate-free version: Let f : M → N is s/a/h. f restricts to an iso p ∈ U → W forsome nbhd U iff (df)p is a linear isomorphism.

Proof of Theorem 7.4:

Analytic case is easy, by formally inverting the power series. Smooth case is harder: useTaylor’s theorem with remainder. �

What you do with it is be clever. We won’t get that far today.

Lemma 7.5: T(p,q)(M ×N) = TpM ⊕TqN , and conversely, i.e. if TpM = V1⊕V2, then there existsa nbhd U 3 p with U ∼= U1 × U2 such that TpU1 = V1 and TpU2 = V2.

23

Proof of Lemma 7.5:

This is obvious in coordinates. The pt-derivations ∂xi = ∂∂xi

are a basis for the tangent space.�

Lemma 7.6: Suppose M ×Nπ

��N

s

YY is a s/a/h section of the projection. Then s is a closed immersion.


**I was a few minutes late.**

Lemma 8.1: Given TpM = V1 ⊕ V2, there is an open nbhd U1 × U2 of p s.t. Vi = TpUi.

Lemma 8.2: If s : N →M ×N is a s/a/h section, then s is a (closed) immersion.

M ×Nπ

��N

s

YY

Proof of Lemma 8.2:

Wolog, M ⊆ Rm (y1, . . . , ym), N ⊆ Rn (x1, . . . , xn). Then s(x1, . . . , xn) = (f1, . . . , fm, x1, . . . , xn),where fi(x1, . . . , xn) is s/a/h. Now change coordinates: ξi = yi − fi(x), xj . Then the imageof s is defined by ξi = 0. �

Proposition 8.3: f : N →M is an immersion on a nbhd of p ∈ N iff (df)p is injective.


The forward direction is obvious. For the reverse, let q = f(p); then (df)p : TpN ↪→ TqM .Let TqM = V1 ⊕ V2 where (df)p : TpN

∼→ V2. Wolog (replacing with smaller nbhds),

Nf //

g

&&LLLLLLLLLLLL M = U1 × U2

π

��U2

g−1

ffLLLLLLLLLLLL

(8.1)

Then s = f ◦ g−1 is a section, so f = s ◦ g is an immersion. �

We will need a bit more about manifolds, because we want to understand how the Lie algebra givesrise to a Lie group. This requires:

24

8.1 Vector Fields

We have a manifold M , and a vector field Xp ∈ TpM : a vector at p for each p ∈M . So at each p,we have a derivation

Xp(fg) = f(p)Xp(g) +Xp(f) g(p) (8.2)

We define (Xf)(p) def= Xp(f). Then X(fg) = f X(g) +X(f) g. So X is a derivation. But it mightbe discontinuous. We define a vector field to be s/a/h if X : SM → SM . In local coordinates, thecomponents of Xp must depend smoothly on p.

Lemma 8.4: The commutator [X,Y ] def= XY − Y X of derivations is a derivation.

Proof of Lemma 8.4:

An easy calculation:

XY (fg) = XY (f) g +X(f)Y (g) + Y (f)X(g) + f XY (g) (8.3)Switch X and Y , and subtract:

[X,Y ](fg) = [X,Y ](f) g + f [X,Y ](g) � (8.4)

We now write down the definition of a Lie algebra. A Lie algebra is a vector space L with a bilinearmap [, ] : L × L → L (i.e. a linear map [, ] : L ⊗ L → L), satisfying

1. Antisymmetry: [X,Y ] + [Y,X] = 0

2. Jacobi: [X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0

It’s a basic fact that every Lie algebra has a faithful representation where [X,Y ] = XY −Y X. Butit’s a deep theorem requiring the structure theory of Lie algebras that every finite-dimensional Liealgebra has a finite-dimensional faithful representation.

Proposition 8.5: [X,Y ] = XY − Y X makes End(V ) into a Lie algebra for any v-space V .

There are other ways, using antisymmetry, to rewrite Jacobi. E.g.:

[X, [Y,Z]] = [[X,Y ], Z] + [Y, [X,Z]] (8.5)

I.e. adX : Y 7→ [X,Y ] is a derivation on L.

[[X,Y ], Z] = [X, [Y, Z]]− [Y, [X,Z]] (8.6)

I.e. ad[X,Y ] = (adX)(adY )−(adY )(adX). I.e. ad : L → End(L) is a Lie algebra homomorphism.Thus, if L has no center (no X such that adX = 0), then ad provides a faithful representation ofL, finite-dimensional if L is.

25

8.2 Integral Curves

Returning to vector fields, let’s integrate. We have a geometric integration of vector fields, asvector-valued functions. Viewing them as derivations and taking their commutators gets awayfrom geometric intuition, so we’d like to present how the commutator of vector fields behavesgeometrically.

Let ∂t be the vector field f 7→ ddtf on R

t. E.g. if γ : I → M where 0 ∈ I ⊆ R is an interval, and

γ(0) = p, then [γ] = dγ0(∂t) is the tangent vector of γ at 0 7→ p.

Proposition 8.6: Given a s/a/h vector field Xp on M and a point p ∈ M , there’s some openinterval I ⊆ R and 0 ∈ I such that there’s a unique s/a/h curve γ : I →M satisfying:

γ(0) = p (8.7)(dγ)t(∂t) = Xγ(t) ∀t ∈ I (8.8)

Question from the audience: Did you mean to give some conditions on I, e.g. maximality?Answer: Well, I can always take a maximal I. It need not be all of R, e.g. if M is simply aninterval in R and Xp = ∂t.


In local coordinates, γ : R → Rn, and we can use existence and uniqueness theorems forsolutions to differential equations; then you need that a s/a/h differential equation has as/a/h solution.

But there’s a subtlety. What if there are two charts, and solutions on each chart, that divergeright where the charts stop overlapping? Well, since M is Hausdorff, if we have two mapsI → M , then their locus where they agree is closed, so if they don’t agree on all of I, thenwe can go to the maximal point where they agree and look locally there. �

We define the integral curve∫X,p(t) of X at p to be the maximal curve satisfying (8.7-8.8)

Remark: The integral depends smoothly on p.

Question from the audience: We gave an example where I cannot be all of R. Suppose M iscompact. Then can we solve differential equations for all time, or are there other obstructions?Answer: I think that on compact manifolds solutions to differential equations extend to R.

Let’s take a point p and two vector fields X and Y . Then we run X and Y alternately, ending atq, and we see that p and q are close by.

•p

•p1

•p2

•p3

•q

∫X

t

::ttttttttt

∫Y

s ��4444

∫X

−t

||yyyyyyyy∫Y

−saaDDDD

26

In quotes, we have

[X,Y ]p = lims,t→0

q − pst

(8.9)

More precisely, for any f , we can show f(q)− f(p) = st[X,Y ]pf +O(s, t)3.

**From Hwajong Yoo:**

Let α(t) =∫X , then f(α(t))′ = Xf(α(t)). Iterating this we get

Äddt

änf(α(t)) = Xnf(α(t)) and

by Taylor series expansion, f(α(t)) =∑ 1

n!

Äddt

änf(α(0))tn =

∑ 1n!X

nf(p)tn = etXf(p). So,f(q) =

Äe−sY f

ä(p3) =

Äe−tXe−sY f

ä(p2) =

ÄesY e−tXe−sY f

ä(p1) =

ÄetXesY e−tXe−sY f

ä(p) and

we already know that etXesY e−tXe−sY = 1 + st[X,Y ]+ higher terms. Hence, we get f(q)− f(p) =st[X,Y ]pf +O(s, t)3.


9.1 Vector fields and group actions

**It seems that today I copy the board, and use e for the identity of a group.**

We continue our story of vector fields on Lie groups. This will be responsible for the fact that theLie algebra really knows a lot about the Lie group.

Let Gy M be a Lie group acting on a manifold. I.e. we have a s/a/h map ρ : G×M → M . LetX ∈ TeG. We attach this to a vector field LX ∈ Vect(M). There are various equivalent ways todo this:

1. Let X = [γ] for some path γ in G. Then we define (LX)m = [γ] where γ(t) = ρ(γ(t)−1,m).Remark: we’re making γ go the wrong way. This vector field acts on functions:

(LX)mfdef=

d

dt

∣∣∣∣t=0

f(γ(t)−1m) (9.1)

Here we really do want the inverse, because if G y M , then G y C(M,R) contravariantly.The problem here is actually much more basic: the problem is that we should have definedthe directional derivative backwards. But try explaining that to Math 53 kids. It’s a lostcause, just like writing functions on the right is a lost cause.

2. Extend X to a vector field X on e ∈ U ⊆ G however you want, and lift this to ˜X on U ×Mby having it point only in the U direction: ˜X(u,m) = (Xu, 0). Trivial check in coordinatesthat this construction is s/a/h. Then LX acts on functions:

(LX)f def= − ˜X(f ◦ ρ)∣∣∣∣{e}×M=M

(9.2)

27

3. Pointwise:(LX)m = −(dρ)e,m(X, 0) (9.3)

Always the same minus sign.

Proposition 9.1: The previous are equivalent.

Proposition 9.2: Given a G-equivariant map f : M → N , then (df)m(LMX) = (LNX)f(m).

Ignore the equivariance for a moment. If we have a map of manifolds, then df is a pointwise thing,taking tangent vectors to tangent vectors. But it does not take vector fields to vector fields, unlessf is an isomorphism.

On the other hand, if Y is any vector field on M , then g ∈ G acts on M isomorphically. Sodg(Y )gm

def= (dg)m(Ym) makes perfect sense. A more natural description: Gy SM by f 7→ f ◦ g−1,an algebra isomorphism (sheafily, it sense f over open U to a s/a/h function on gU). So if we thinkof Y : SM → SM as a derivation, then gY g−1 is a derivation. This is a very natural way to saythat G acts on vector fields. We define gY

def= gY g−1. Then clearly gY = dg(Y ).

Ok, so let’s apply the proposition to g : M → M by some group element g ∈ G. But this isn’t anequivariant map: g may not commute with the rest of the group. We can fix this. Let gM be Mwith an adjusted action. If ρ : GyM was our old action, then

gρ(h,m) = h ·gm = ρ(ghg−1,m) (9.4)

Then Mg→ gM sending m 7→ gm is G-equivariant:

m � g //_

h��

gm_gh

��hm

� g // ghm = ghg−1gm

Corollary 9.2.1: gLX = dg(LX) = LgMX = L(Ad(g)X) where if X = [γ], then Ad(g)X =

[g γ(t) g−1], i.e. Ad(g) = d(g − g−1)e.

Ad is the only natural action Gy TeG.

Our convention is that actions are always left. When we say “right action”, we mean the actionfrom the right by in the inverse of a group element.

Proposition 9.3: For the right action of G on itself, L : TeG → Vect(G) is an isomorphism ofTeG on left-invariant vector fields. Also, (LX)e = X.

If you use the other hands, then you get (LX)e = −X.


The point is that the left- and right-actions commute. Let λg : G → G be h 7→ gh. This isequivariant for the right-action. So dλg(LX) =λg (LX) = LX, so LX is left-invariant. Also

28

(LX)e = X becuase ρ(g, e) = g−1. But if our field is invariant, knowing is at a point tells useverything:

(LX)g = (dλg)e(LXe) = (dλg)e(X) �

Well, Vect(M) is a Lie algebra, and if GyM , and hence on the function space, then the invariantderivations are a Lie subalgebra. So, we define the Lie algebra of a group G to be Lie(G) def= TeG,with the bracket induces by the commutator of left-invariant vector fields.

Now, maybe you should worry: are there two Lie algebras, e.g. the one where we switch hands?Well, yes, but it’s just the opposite Lie algebra: the right-bracket is minus the left-bracket. Thiswill be an exercise.

We had better check some things.

Lemma 9.4: Given s/a/h G y M , and X ∈ Lie(G) represented by X = [γ], and Y ∈ Vect(M),then

d

dt

∣∣∣∣t=0

γ(t)Y f = [LX, Y ]f (9.5)

We should think of Vect(M) as a manifold; this tells us the action of Lie(G) y Vect(M).

Proof of Lemma 9.4:

d

dt

∣∣∣∣t=0

γ(t)Y f (p) =d

dt

∣∣∣∣t=0

γ(t)Y γ(t)−1f (p) (9.6)

=d

dt

∣∣∣∣t=0

γ(t)Y f(γ(t) p) (9.7)

=d

dt

∣∣∣∣t=0

γ(t)Y f(γ(0) p) + γ(0)d

dt

∣∣∣∣t=0

Y f(γ(t)p) (9.8)

= LX(Y f)(p) + Yd

dt

∣∣∣∣t=0

f(γ(t)p) (9.9)

= LX(Y f)(p) + Y (−LX f)(p) (9.10)= [LX, Y ]f (p) � (9.11)

Corollary 9.4.1: Same setup: GyM . If X,Y ∈ Lie(G), where X = [γ], then

LMÄLAd(−X)Y

ä=

d

dt

∣∣∣∣t=0

L (Ad (γ(t))Y ) f = [LX,LY ] f (9.12)

Consider M = Gx G. Then what we’re saying is that

−LAd(X)Y = [X,Y ] def= (adX)Y (9.13)

29

Then for any GyM , thenL ([X,Y ]) = [LX,LY ] (9.14)

This is the right way to think of the Lie bracket on the tangent space. It’s the universal bracketthat becomes the commutator for any infinitesimal left action.


We begin by reminding where we are from last time.

To any Lie group, we can induce a Lie bracket on its tangent space at the identity: we extend eachtangent vector uniquely to a left-invariant vector field, and take the Lie bracket of vector fields.But more importantly, this is a universal construction: the infinitesimal action on the right on amanifold commutes with any left action.

In Lie(G), using the fact that G acts on its tangent-space at the identity via the adjoint action, wecan directly define the bracket:

[X,Y ] = −LAdX(Y ) (10.1)

=d

dt

∣∣∣∣t=0

Ad(γ(t))Y (10.2)

where X = [γ].

We have already defined the Lie bracket for closed linear groups. We should verify that our newnotion matches the old one.

Lemma 10.1: The abstract Lie bracket on Lie(GLn) = gln = TeGLn = Mn is the matrix bracket[X,Y ] = XY − Y X.

Proof of Lemma 10.1:

We simply compute the formula. If X ∈ gln, then ddt

∣∣∣t=0

etX = X. The adjoint action isAdG(g)h = ghg−1.

[X,Y ] =d

dt

∣∣∣∣t=0

etXY e−tY = XY − Y X (10.3)

Question from the audience: Do you mean the adjoint action on G or on Lie(G)? An-

swer: Well, Gy G by g · h = ghg−1, but this fixes e and is linear in h, so GLnAdyMn = gln

by Y 7→ gY g−1. �

Now, if H ↪→ GLn is a closed linear group, then Lie(H) ↪→ Lie(G).

Corollary 10.1.1: The same holds for closed subgroups H ⊆ GLn.

We should clarify something about integral curves. Let X on M 3 p be a vector field, and thendefine

Ä∫pXä

(t) to be a curve γ(t) : I → M where I ⊆ R is an interval, satisfying γ(0) = p and

30

dγ(∂t) = Xγ(t). Then everything works out: the manifold is Hausdorff, and we have existence anduniqueness theorems.

What if we are in the complex case? Well, we could just treat complex manifolds as twice-dimensional real manifolds. But holomorphic differential equations have uniqueness and existencetheorems, so we can talk about complex integral curves. We should defined

Ä∫pXä

to be a functionU →M where 0 ∈ U ⊆

openC.

10.1 The exponential map

We saw that the Lie algebra knows about e.g. homomorphisms of its group, via the exponentialmap. We should generalize this.

We now state a special case of the general statement that Lie subalgebras should give us Liesubgroups.

Lemma 10.2: For X ∈ Lie(G) there is a unique Lie group homo γX : R → G (or C → G in thecomplex case) — C and R have obvious Lie group structures — such that (dγX)0(∂t) = X.It’s given by γX(t) =

(∫e LX

)(t).


There is an integral curve defined on some neighborhood of 0 ∈ R (C, but we’ll use the realpicture). Let γ : I → G be this integral curve. Well, LX is left-invariant; gγ(t) then is anintegral curve through g. So let g = γ(s) for s ∈ I. So this moves the curve over. But bothγ(t) and γ(s)γ(t) are integral curves through s, so they must be the same on the overlap:γ(s+ t) = γ(s)γ(t). (Also γ(−s) = γ(s)−1 for s ∈ I ∩ (−I).) So this integral curve is a pieceof a group homomorphism. Then we can extend the homomorphism to I + I; this extensionis well-defined because γ is a partial homomorphism. Since R is archimedean, we can iterateto extend this to all of R (or C, by doubling the open neighborhood again and again). And itwill continue to satisfy the group homomorphism rule, and conversely it must be an integralcurve through any point because it is at the identity. �

Corollary 10.2.1: There is a bijection between 1-parameter subgroups of G (homos R→ G) andelements of the Lie algebra.

Then 10.2 gives us a map exp : Lie(G) → G by expX def= γX(1). Then exp tX = γX(t). It’s clearthat exp tX is a s/a/h function of t. But we should show that it’s smooth as a function of X.

Proposition 10.3: Let X(b) be a s/a/h family of vector fields on M . We should define this: The“vertical” vector field on B×M given by X(b,m) = (0, (X(b))m) is s/a/h. Question from theaudience: What is B? Answer: Our family is parameterized by b ∈ B. So in particular ifyou go back to look at how we defined LX: when B = Lie(G) and G y M then X(b) = Lbis a smooth family.

31

Refresh: Let X(b) be a s/a/h family of vector fields parameterized by b ∈ B a manifold, thenÄ∫pX

(b)ä

(t) is a s/a/g map B ×M × R→M . (Or R C.)

The proof is shorter than the statement:


Note that Ç∫(b,p)

X

å(t) =

Çb,

Ç∫pX(b)

å(t)å

(10.4)

So B ×M ×R→ B ×M π→M by (b, p, t) 7→Ä∫

(b,p) Xä

(t) 7→Ä∫pX

(b)ä

(t) is a composition ofs/a/h functions, hence is s/a/h. �

Theorem 10.4: There is a unique s/a/h map exp : Lie(G) → G such that for each X ∈ Lie(G),t 7→ exp(tX) is the integral curve of LX through e, and is a Lie group homo (R or C)→ G.

We are not saying that exp is a Lie group homo; Lie(G) is a Lie group under +, but G is notcommutative. It is a Lie group homo along any line.

Proposition 10.5: The differential at the origin (d exp)0 is the identity map idLieG.


d(exp tX)0(∂t) = X. �

Corollary 10.5.1: exp is an isomorphism of a neighborhood of 0 ∈ Lie(G) onto a neighborhoodof e ∈ G. I.e. exp is a local homeomorphism. We call its (local) inverse “log”.

Proof of Corollary 10.5.1:

Its derivative is invertible. �

Corollary 10.5.2: If G is connected, then exp(Lie(G)) generates G.


Any open nbhd generates the connected component. **Why again?** �

Theorem 10.6: Given φ : H → G a Lie group homo, then we have a commutative diagram:

Hφ // G

Lie(H)

exp

OO

(dφ)e // Lie(G)

exp

OO (10.5)

Corollary 10.6.1: If H is connected, then (dφ)e determines φ.

Proposition 10.7: exp : gln → GLn is matrix exponential.

32


Suffice to show that exp defines an integral curve. But t 7→ etx is a s/a/h group homomor-phism, and d

dt

∣∣∣t=0

etX = X. �

Corollary 10.7.1: Same statement for closed H ⊆ GLn.


Use 10.6. �

It would be nice if any Lie algebra map induced a Lie group map, but this is of course not true:

Example: R → R/Z = U(1) induces the identity on Lie algebras. But the inverse of the map onLie algebras does not induce a map R/Z→ R, because of a topological obstruction.

It turns out that that’s the whole issue. If you have a Lie algebra homomorphism and the corre-sponding groups are simply connected, then there is a group homomorphism.

There is a weak form:

Fact: Given ψ : Lie(H)→ Lie(G), we can find ψ = expG ◦ψ ◦ logH . By commutativity in 10.6, wehave ψ(xy) = ψ(x) ψ(y) etc., provided it makes sense. I.e. ψ is a partial group homomorphism.

The right theorem is an equivalence of categories between Lie algebras and simply connected Liegroups.


In the next few lectures we will build up to building the basic theorem relating Lie algebras a Liegroups.

Theorem 11.1: Fundamental Theorem on Lie Groups and Algebras

(a) We saw that homomorphisms from Lie algebras do not always lift to Lie group homo-morphisms, e.g. U(1)→ R. But this failure is topological:

The functor G 7→ Lie(G) gives an equivalence of categories between

Simply Connected Lie Groups/R or C↔ Lie Algebras/R or C

(b) You have to do some choosing, because a Lie algebra does not determine its group. But:

“The” inverse functor h 7→ Grp(h) is left-adjoint to G 7→ Lie(G).

**The functor Grp lands in Simply-connected Groups, but the adjunction isbetween Lie Groups and Lie Algebras.**

33

The standard proof is not how we’re about to sketch, but we’ll say some words about how oneought to prove such a theorem.

Idea of proof of Theorem 11.1:Lie(G) G

⊆ ⊆

Uexp //

Vlog

oo

∈ ∈

0 e

We have V ×V µ→ G a partial multiplication; we should think of this as (V ×V )∩µ−1(V )→ V .So we should define a “partial groups law” B : open→ U , where open ⊆ U × U , via

B(X,Y ) = log(expX expY ) (11.1)

Then should show that the Lie algebra operation determines B.

The hard part, which we may not do: Given h, define B, and build H as the group freelygenerated by U mod the relations XY = B(X,Y ) if X,Y,B(X,Y ) ∈ U . The problem is thatif we do this with an abstract group, it’s not clear it’s a Lie group at all, or simply connected.So we will need:

Claim: H is a Lie group, with U as an open submanifold. �

Corollary 11.1.1: Every Lie subalgebra h of Lie(G) is Lie(H) for a unique connected subgroupH ↪→ G, up to equivalence.

This will also follow from some general remarks about covering spaces. We will need the correctnotion of “subgroup”, so that R ↪→ T 2 irrationally is a subgroup.

Sketch of standard proof of 11.1: First prove the corollary. Then use:

Theorem 11.2: Ado’s Theorem

Every finite-dimensional lie algebra h is isomorphic to a subalgebra of gln.

This is strong extra input, so it makes it less satisfactory to think about.

No matter what you do, we’ll need this operation B, so let’s begin by talking about that.

Question from the audience: So in Ado’s theorem, you mentioned “finite dimensional”. Shouldthat be in the theorem too? Answer: When we say “Lie group”, we mean in particular a finitedimensional manifold. So, yes, we should have finite-dimensional Lie algebras, too.

Theorem 11.3:

(a) The formal power series B(tX, sY ) = log(etXesY ) in T (X,Y )[[s, t]], where T (X,Y ) isthe non-commutative free “tensor” algebra generated by X and Y , and exp and log aredefined formally as always — B is given by the formula

B(tX, sY ) = tX + sY + st12

[X,Y ]− st2 112

[X, [X,Y ]]− s2t112

[Y, [Y,X]] + . . . (11.2)

34

— has coefficients that are all Lie-bracket polynomials in X and Y .

(We will give a conceptual proof, although once you know the result, it’s not hard tosolve for the coefficients. Question from the audience: If what you’ve written downare “Lie-bracket monomials”, what’s a “polynomial”? Answer: Linear combinations ofmonomials.)

(b) Given a Lie group G, there exists a neighborhood U ′ 3 0 in Lie(G) such that U ′ ⊆ Uexp

�log

V ⊆ G and B(X,Y ) converges on U ′ × U ′ to log(expX expY ).

We have removed the ss and ts in the second part. This is because we didn’t want to discuss formalpower series in noncommuting variables. We don’t want to talk about the completion of the tensoralgebra.

Proof of 11.3 part (b):

We begin with a basis identity: exp(tX) is an integral curve to LX through e. So by left-invariance, t 7→ g exp(tX) is the

∫curve to LX thorugh g. Thus, for f analytic on G,

d

dtf(g exp tX) = (LX · f)(g exp tX) (11.3)

We will begin abusing notation, writin X for LX. Thus, by iterating,

d

dt

n

f(g exp tX) = (Xnf)(g exp tX) (11.4)

If f is analytic, then for small t the Taylor series converges:

f(g exp tX) =∑

(Xnf) (g)tn

n!= (etXf)(g) (11.5)

Doing it again,

f(exp tX exp sY ) =Å∑

Y n sn

n!f

ã(exp tX) =

Å∑XmY n s

ntm

n!m!f

ã(e) = (etXesY f)(e)

(11.6)Now, f could be anything. For example, a coordinate function, or a coordinate of log. Inparticular,

ÄetXf

ä(e) = f(exp tX), thought of as exp tX ∈ G. So we concludeÄ

eB(tX,sY )fä

(e) =ÄetXesY f

ä(e) (11.7)

Thus (b) follows from (a). We could have done this via differential equations, which wouldwork in C∞ rather than analytic.

The cleanest way to do (a) requires extra machinery. So we postpone a bit. �

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11.1 Universal Enveloping Algebras

A representation of a Lie group is a homomorphism G→ GL(n,R) (or C). Thus, a representationof Lie algebras is a homomorphism Lie(G) → gln = End(V ), which is a Lie algebra given by[X,Y ] = XY − Y X. We want an associative algebra U(g) such that the Lie algebra reps of g areexactly the U(g) modules. This is the universal enveloping algebra:

U(g) = T (g)/ 〈xy − yx− [x, y] : ∀x, y ∈ g〉 (11.8)


In the last three minutes, we mentioned the concept of the Universal Enveloping Algebra of a Liealgebra. We will use this to prove the Campbell-Hausdorff-Baker **perhaps?** formula. Also,there were some objections to the convergence.

12.1 The Universal Enveloping Algebra

The basic concept of a tensor algebra over a vector space V is that you take a basis of V , and thentake the algebra of noncommuting polynomials in the basis:

T (V ) =⊕n≥0

V ⊗n (12.1)

with multiplication given by ⊗ : V ⊗k × V ⊗l → V ⊗(k+l). This is the free associative algebragenerated by V , i.e. any a linear map V → A a (possibly noncommutative) K-algebra extends to aunique algebra homomorphism T (V )→ A. This is the best way of doing it, because it’s the mostuniversal. **So T (·) is adjoint to Forget : alg→ vect.**

Given a Lie algebra g, we define the universal enveloping algebra to be

U(g) = T (g)/〈[x, y]− (xy − yx)〉 (12.2)

This satisfies a universal property: we can regard any associative algebra A as a Lie algebra with[a, b] = ab− ba. Then any Lie algebra homo φ : g→ A “extends” uniquely to an associative algebrahomomorphism U(g) → A. The word “extends” is a little funny, because a priori g might notembed in U(g). It does, in fact, but this is not obvious. What we can say is that U(g) comes witha map g→ U(g), and we want φ to factor through this map.

In categorical language, g 7→ U(g) is a functor by the universal property or directly from theconstruction. It’s left-adjoint to Forget : associative algebras→ lie algebras.

One reason you should want such a thing — there are a number of reasons that you should wantsuch a thing, and I can name two or three off the top of my head. We wrote down some axioms thatmay not completely characterize the brackets we want. But the fact that Lie algebras embed in

36

their universal enveloping algebras says that Jacobi is enough to characterize a Lie bracket. Also,U will help us prove the CHB formula. And we want to study representations.

We define a representation (or “g-module”) V of a Lie algbera g to be a Lie algebra homomorphismg→ End(V ). But by the universal property, g−mod = U(g)−mod.

Example: If g is an abelian Lie algebra, i.e. [g, g] = 0, then U(g) = S(g) is the symmetric algebra.One should expect the noncommutative case should be some deformation of this.

Example: If f is the free Lie algebra on generators x1, . . . , xd — we had better make this precise,because it’s more general then the tensor algebra, because we’re taking a vector space witha basis of all bracketed words — so we take non-associative words, and then mod out by theanti-symmetry and Jacobi identities. Then any map {x1, . . . , xd} → g extends to a uniqueLie alg homo f → g. There’s always a notion of “Free”. **Free is adjoint to Forget toset.** Then by tracing universal properties, U(f) is the tensor algebra T (x1, . . . , xd). So ifyou believe that the free Lie algebra embeds in its universal enveloping algebra, then the freeLie algebra can be constructed from the tensor algebra.

The tensor algebra is graded, but the relations in U(g) = T (g)/〈xy − yx = [x, y]〉 are not homoge-neous. But U(g) does have a filtration, where the nth part of U is the image of the degree-at-most-npart of T :

U(g)≤0def= K

U(g)≤1def= K+ g

U(g)≤ndef= (U(g)≤1)n

In the second line, g is really the image of g in U(g). Then

U(g)≤k U(g)≤l ⊆ U(g)≤k+l (12.3)

grU(g) =⊕k

U(g)≤k/U(g)<k (12.4)

is a graded associative algebra generated by the image of g. But it’s commutative: if x, y ∈ g andx, y ∈ gr1 U(g), then xy − yx = [x, y] = 0 ∈ gr2 U(g). I.e. S(g)� grU(g).

Theorem 12.1: (Poincare-Birkhoff-Witt)

S(g)→ grU(g) is an isomorphism.

Corollary 12.1.1: g ↪→ U(g), i.e. every Lie algebra is isomorphic to a subalgebra of some End(V ),e.g. V = U(g) as a left U(g)-module.

This justifies the Jacobi identity.

Question from the audience: Could you clarify how g→ U(g)? Answer: Well, K⊕g→ U(g)≤1,and the grading is simple enough at this level.

37

We will reformulate the Poincare-Birkhoff-Witt theorem. Take a basis of S(g). In particular, let’stake an ordered basis xα of g. Then let’s think of each monomial in S(g) in order, as a productxα1 . . . xαn where α1 ≤ · · · ≤ αn. This is a basis of the symmetric algebra. Well, the constructionS(g)→ U(g) is an algebra homomorphism, so xα1 . . . xαn 7→ xα1 . . . xαn ∈ U(g)≤n/U(g)<n. They’regoing to span, so we want to show that they’re independent. It’s enough to show that they’reindependent in U(g)≤n.

Remark: PBW (12.1) is equivalent to the statement that the monomials {xα1 . . . xαn : α1 ≤ . . . αn}are independent of U(g).

Beginning of proof of 12.1:

Choose a basis of g that’s well-ordered. We want to show that the basis is independent inU(g) = T (g)/I, so we want to show that no linear combination is in I. So let J be theK-linear span of expressions in T (g) of the form:

xα1 . . . xαn (xβxγ − xγxβ − [xβ, xγ ])xν1 . . . xνl (12.5)

such that α1 ≤ · · · ≤ αn ≤ β > γ. The leading term is x~αxβxγx~ν in the degree-lexicographicorder on T (g). It’s clear that J ⊆ I, and that J is a right-ideal. Moreover, J containsexpressions that generate I as a two-sided ideal. So suffice to show that J is a left-ideal, sincethen J = I and the things in I have a natural order.


Last time, we introduced the

PBW Theorem: S(g) ∼→ grU(g).

Proof of PBW Theorem:

We’ve reduced last time the problem to showing that given an ordered basis (xα) of g, wewant to show that S def= {xα1 . . . xαn : α1 ≤ . . . αn} is independent in U(g).

So let U(g) = T (g)/I, and define J ⊆ T (g) to be the span of expressions

X = xα1 . . . xαk (xβxγ − xγxβ − [xβ, xγ ])xν1 . . . xνl (13.1)

where α1 ≤ · · · ≤ α)k ≤ β > γ. We take the deg-lex orderer of monomials in T . The leadingmonomial in 13.1 is x~αxβxγx~ν . Thus, S is independent in T (g)/J . This is some sort ofnoncommutative Groebner basis.

Well, I is generated by expressions of the form xβxγ − xγxβ − [xβ, xγ ] as a left-right ideal.If β > γ, then the expression is in J . If β < γ, then switch them and still in J , usingantisymmetry: [X,Y ] = −[Y,X]. If β = γ, then it’s 0.

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Side remark: [X,X] = 0 from antisymmetry, if the characteristic is not 2. But the betterantisymmetry axiom is exactly that [X,X] = 0 ∀X. This and bilinearity imply that [X,Y ] +[Y,X] = 0.

Anyway, J is a right ideal contained in I, and the left-right ideal generated by J contains I.So suffice to show that J is a left-ideal.

So, we multiply xδX. If k > 0 and δ ≤ α1, then xδX ∈ J . If δ > α1, then xδX ≡xα1xδxα2 · · · + [xδ, xα1 ]xα2 . . . mod J . And both xδxα2 . . . and [xδ, xα1 ]xα2 . . . are in J byinduction on degree. Then since α1 < δ, xα1xδxα2 · · · ∈ J by induction on δ. We’re doing agross transfinite induction here, on degree and each index.

So suffice to show that if k = 0, then we’re still in J . I.e. if α > β > γ, then we want to showthat xα (xβxγ − xγxβ − [xβ, xγ ]) ∈ J . Well, since α > β, we see that xαxβ−xβ− [xα, xβ] ∈ J ,and same with β ↔ γ. So, working modulo J , we have

xα (xβxγ − xγxβ − [xβ, xγ ]) ≡ (xβxα + [xα, xβ])xγ − (xγxα + [xα, xγ ])xβ − xα[xβ, xγ ]≡ xβ (xγxα + [xα, xγ ]) + [xα, xβ]xγ − xγ (xβxα + [xα, xβ])− [xα, xγ ]xβ − xα[xβ, xγ ]

≡ xγxβxα + [xβ, xγ ]xα + xβ[xα, xγ ] + [xα, xβ]xγ − xγ (xβxα + [xα, xβ])− [xα, xγ ]xβ − xα[xβ, xγ ]

= [xβ, xγ ]xα + xβ[xα, xγ ] + [xα, xβ]xγ − xγ [xα, xβ]− [xα, xγ ]xβ − xα[xβ, xγ ]≡ −[xα, [xβ, xγ ]] + [xβ, [xα, xγ ]]− [xγ , [xα, xβ]]= 0 by Jacobi. �

13.1 U(g) is a bialgebra

An algebra over K is a **monoid object in VectK** vector space U along with a linear mapµ : U ⊗

KU → U which is associative, i.e. it satisfies

U ⊗ U ⊗ Uµ⊗1U

xxppppppppppp1U⊗µ

&&NNNNNNNNNNN

U ⊗ U

µ''NNNNNNNNNNNN ◦ U ⊗ U

µwwpppppppppppp

U

(13.2)

**and identities**.

A coalgebra is an algebra in Vectop. I.e. it is a space V along with a map ∆ : V → V ⊗ Vsuch that the reverse diagram commutes. In elements, if ∆x =

∑x(1) ⊗ x(2), then we want∑

x(1) ⊗∆(x(2)) =∑

∆(x(1))⊗ x(2).

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A bialgebra is an algebra and a coalgebra, with some coherence. In particular, ∆ is an algebra ho-momorphism (or equivalently µ is a coalgebra homomorphism). **So a bialgebra is a coalgebraobject in Algebras?**

Proposition 13.1: U(g) is a bialgebra with ∆ : U(g)→ U(g)⊗U(g) such that ∆(x) = x⊗1+1⊗xfor x ∈ g. We say that x is primitive if ∆(x) = x⊗1+1⊗x, and write primU for the primitiveelements of a bialgebra U .


To construct ∆ : g→ U ⊗ U , we need to show that

x 7→ x⊗ 1 + 1⊗ x ∈ U ⊗ U

is a Lie algebra homomorphism, because an algebra homomorphism from U is exactly a Liehomomorphism from g. But

[x⊗ 1 + 1⊗ x, y ⊗ 1 + 1⊗ y]U⊗U = [x⊗ 1, y ⊗ 1] + [1⊗ x, 1⊗ y]= [x, y]⊗ 1 + 1⊗ [x, y]

Moreover, ∆ is coassociative on a basis: ∆2(x) = x⊗ 1⊗ 1 + 1⊗ x⊗ 1 + 1⊗ 1⊗ x. �

Corollary to PBW: g = prim (U(g))

Proof of Corollary to PBW:

Filter U(g) ⊗ U(g) in the obvious way. Since ∆ is an algebra homomorphism, ∆U(g)≤1 ⊆(U(g)⊗ U(g))≤1, so ∆(U(g)≤n) ⊆ (U(g)⊗ U(g))≤n, so ∆ induces ∆ on grU(g) = S(g).

If ξ ∈ U(g)≤n is primitive, then ξ ∈ grn U(g) is primitive. In a polynomial ring S(g)⊗S(g) =K[yα, zα], where {xα} are a basis, and yα = xα⊗ 1, zα = 1⊗xα. But f ∈ S(g) is primitive ifff(y + z) = f(y) + f(z), i.e. f is homogeneous of degree 1. Thus, every primitive ξ ∈ U(g) issome x+ c, where x ∈ g and c ∈ K. But x is primitive, so c is primitive, and a constant canonly be primitive if it’s 0. �

**We never supplied a formal definition of filtered algebra, although we used thenotion. Wikipedia includes that

A filtered algebra over the field K is an algebra (A, ·) over K which has anincreasing sequence {0} ⊆ F0 ⊆ F1 ⊆ · · · ⊆ Fi ⊆ · · · ⊆ A of subspaces of A suchthat

A = ∪i∈NFi

and that is compatible with the multiplication in the following sense

∀m,n ∈ N, Fm · Fn ⊂ Fn+m.

(I’ve changed the notation slightly.) To each filtered algebra A, we construct anassociated graded algebra grA, and we again quote Wikipedia with minimal notationchanges:

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As a vector space gr(A) =⊕

n∈NGn , where, G0 = F0 , and ∀n > 0, Gn =Fn/Fn−1. The multiplication is defined by (x+ Fn)(y + Fm) = x · y + Fn+m+1

We can extend this notion to get, e.g., a filtered module and the associated gradedmodule.

In all cases, we should make the definitions categorical. We have a category FilVectof filtered vector spaces, whose objects are vector-spaces along with filtrations intosubspaces as above, and whose morphisms respect the filtration: if φ : V → W , thenφ(Vn) ⊆Wn. The tensor-product also respects the filtration: (V ⊗W )n =

∑k+l=n Vk⊗Wl.

Then filtered algebras and filtered modules are just algebra- and module-objects inFilVect. Any filtered vector space has an associated graded vector space, and this is amonoidal functor from FilVect→ GrVect. **


We’re still moving towards the Baker-Campbell-Hausdorff formula. First, we will talk about thegeometric interpretation of the Universal Enveloping Algebra.

Suppose U ⊆open

M a manifold, and consider ζ(u) ∈ S(U) a s/a/h function. Then we have operators

S(U)→ S(U):

• f ∈ S(U) acts as h 7→ fh.

• X ∈ Vect(U) acts as a derivation.

We can take the bracket of operators. Functions commute, vector fields we understand, and[X, f ]g = X(fg) − fX(g) = X(f) g, so [X, f ] = X(f) ∈ S(U) y S(U). We define a non-commutative algebra D(U) of differential operators to be generated by S(U) and Vect(U). ThenS(U) ⊕ Vect(U) is a Lie subalgebra. Grothendieck gave a more abstract definition: an nth-orderdifferential operator is something whose commutator with a function is a n − 1th order operator.Question from the audience: So [X, f ] ∈ S(U) if and only if X ∈ Vect(U)? Answer: Yes. Itmust be a derivation for that to hold.

Every nth-order differential operator is fXα1 . . . Xαn . If X1, . . . , Xd are a basis of Tu(M) for eachu ∈ U , then D(U) =

∑α1≤···≤αn S(U)Xα1 . . . Xαn . E.g. given coordinates x1, . . . , xd we have vector

fields ∂x1 , . . . , ∂xd , which work like this. So expressions of the form∑~α f~α∂x~α . The general case is

the same: the change-of-basis matrix from 〈X1, . . . , Xd〉 ↔ 〈∂x1 , . . . ∂xd〉 is an invertible matrix offunctions.

So, let’s let U = M = G be a Lie group, and talk about the left-invariant differential operatorsD(G)G. Well, the left-invariant functions are just the constants, because the group acts transitively.If X1, . . . , Xd are left-invariant — i.e. a basis of g = Lie(G) — then f~α∂x~α is left-invariant only iff is. So there’s a natural map U(G)→ D(G)G, but in fact it’s iso: the ordered monomials in U(g)go to a basis, so in particular we have a cheep geometric proof of PBW when g = Lie(G).

41

14.1 Baker-Campbell-Hausdorff formula

There’s some disagreement about the order of the names; most English-language books use “CBH”.

Theorem 14.1: (a) We have an identity in T (X,Y )[[s, t]]:

etXesY = eB(tX,sY )

where B(tX, sY ) = tX + sY + 12st[X,Y ] − 1

12st2[X, [X,Y ]] − 1

12s2t[Y, [Y,X]] + . . . is a

series with coefficients in the free Lie algebra on two generators X and Y **which ofcourse is a quotient of the free tensor algebra T (X,Y )**.

(b) If G is a Lie group, then there are neighborhoods 0 ∈ U ′ ⊆open

U ⊆open

Lie(G) = g and

0 ∈ V ′ ⊆open

V ⊆open

G such that Uexp

�log

V and U ′ � V ′ and B(X,Y ) converges on U ′ × U ′

to log(expX expY ).


We don’t really need the s and t in part (a) if we’re willing to work with formal power seriesin non-commuting variables. But we would also like Taylor series in commuting variables.Anyway, let f be the free Lie algebra on X and Y ; then T (X,Y ) = U(f). We can extend∆ : U(f) → U(f) ⊗ U(f0 to ∆U(f)[[s, t]] → (U(f) ⊗ U(f))[[s, t]] linearally, which is an (s, t)-adically continuous algebra homomorphism.

Lemma 14.2: If U is any bialgebra, and ψ ∈ U [[s, t]] with ψ(0, 0) = 0, then ψ is primitiveterm-by-term — i.e. ∆ψ = ψ ⊗ 1 + 1 ⊗ ψ — if and only if eψ is “group-like” — i.e.∆eψ = eψ ⊗ eψ.

We’re using U [[s, t]]⊗ U [[s, t]]→ (U ⊗ U)[[s, t]] a bialgebra homomorphism.


eψ ⊗ eψ = (1⊗ eψ)(eψ ⊗ 1)

= e1⊗ψeψ⊗1

= e1⊗ψ+ψ⊗1 �

Well, etXetY is grouplike: ∆(etXesY ) = ∆etX ∆esY = (etX ⊗ etX)(esY ⊗ esY ) = etXesY ⊗etXesY . Therefore B(tX, sY ) is primitive term-by-term. That does part (a).

**Charley later asked: Why does log commute with ∆? Because ∆ is an alge-bra homomorphism — commutes with polynomials — and ∆ is continuous —commutes with infinite sums. So ∆ commutes with all formal power series.**

For part (b), we have the multiplication µ : G × G → G. Set up U ′, U, V ′, V **by findingU, V and V ′ so that µ : V ′ × V ′ → V **. Let’s define β(X,Y ) = log(expX expY ), which isa/h as a function of X and Y **∈ U ′**. (We don’t want smooth, because we will use Taylor

42

series.) We have g = Lie(G) = left-invariant vector fields. Then U(g) ∼→ left-inv differentialoperators on G.

Lemma 14.3: If D ∈ U(g) satisfies Df (e) = 0 for all f ∈ S(G)e, then D = 0.


For g ∈ G, Df(g) = λg−1 = D(λg−1f)(e) = 0. �

So, given X,Y ∈ g, f ∈ S(G)e, then (etXetY f)(e) is a Taylor series of f(exp tX exp sY ). Wesaw this before: (Xf)(exp tX) = d

dtf(exp tX). Let’s think of eZf(e) as a formal power seriesin coordinates of Z ∈ g. Then eZf(e) is the Taylor series of f(expZ).

Ok, so let β be the formal series which is the Taylor series of β. Then eβ(tX,sY )f(e) is alsothe Taylor series of f(exp tX exp sY ). This implies that etXesY f(e) and eβ(tX,sY )f(e) havethe same coefficients for every f , but those coefficients are left-inv differential operators, soby the lemma these two series are identically equal. So the formal series of the logs are thesame:

B(tX, sY ) = β(tX, sY )

But β is a/h, so its series converges on a neighborhood U ′′ ⊆open

U ′. By shrinking U ′ and V ′,

we complete the proof. �

We remark that we didn’t really need U(g) = left-inv differential operators, just that two left-invdifferential operators are the same if they’re equal at the identity.

For the last five minutes, we’ll outline where we’re going.

Our goal: Given G and h ≤ Lie(G) a Lie subalgebra, we want to show that h = Lie(H) for some Liesubgroup H ≤ G. What’s a Lie subgroup? Well, it’s a subgroup, with its own manifold structure(not necessarily inherited) so that the inclusion H ↪→ G is s/a/h. **Is a local immersion?** Hshould be unique if it’s connected.

How will we prove this? The exp and log maps take h∩U to an immersed submanifold of V ⊆open

G.

So we want to let H be the group generated by exp(h∩U). But we’ll need to control this carefully.Well, B(X,Y ) converges to h ∩ V if X,Y in a small enough neighborhood of 0 ∈ h. We’ll needmore theorems that something that acts locally like a group really is part of a group.

Lecture 15 October 1, 2008

There are more problems on the website. As always, part of the homework is to figure out whichstatements are false.

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15.1 Subgroups

A Lie subgroup H ≤ G is a subgroup, which is also a Lie group (but not necessarily the inducedtopology) such that the inclusion H ↪→ G is a local immersion. Then, yes, the structure of H as amanifold is determined by G, but in a local way.

Theorem 15.1: Every Lie subalgebra of Lie(G) is Lie(H) for a unique connected Lie subgroupH ≤ G.

Of course, if H ≤ G, then TeH ≤ TeG, so we can think of Lie(H) ≤ Lie(G).

The proof is in two parts: BCH and a bit of topological group theory.


Uniqueness: if H is any Lie subgroup of G, with h = Lie(H) and g = Lie(G), then

H� � // G

h

exp

OO

� � // g

exp

OO

This shows that expG(h) ⊆ H, and if H is connected expH(h) = expG(h) generates H. So His unique as a group. But it’s manifold structure is also given:

0 ∈ U ⊆ g

e ∈ V ⊆ G��

exp log

OO

And exp(U ∩ h) ∼→logU ∩ h is an immersion, giving a chart around e ∈ H, which pushes to any

other point to determine uniquely the topology and manifold structures.

Existence: We have0 ∈W ⊆ g

e ∈ V ⊆ G��

exp log

OO

We choose smaller W ′ ↔ V ′ such that

• (V ′)2 ⊆ V

• B(X,Y ) converges on W ′ ×W ′ to log(expX expY ).

• hV ′h−1 ⊆ V for h ∈ V ′.

Gg 7→hgh−1

// G

g

exp

OO

Ad(h) // g

exp

OO

44

• If h = expX, then Ad(expX) = exp(adX) in some nbhd of 0 ∈ g, so we demand thatexp(adX)Y converges on W ′ ×W ′ to (expX)(expY )(expX)−1.

• B(X,Y ), (exp adX)Y ∈ h ∩W for X,Y ∈ h ∩W ′.

• (V ′)−1 = V ′.

Question from the audience: By exp ad, what do you mean? Answer: (exp adX) def=∑(adX)n/n! as a linear operator. Oh, so it’s always true that Ad(expX) = exp(adX), so

we don’t need the fourth one.

Question from the audience: That fourth one doesn’t even make sense? Answer: No,it should be more like eadXY → log

((expX)(expY )(expX)−1

), where et is a formal power

series.

Ok, so define H ≤ G to be the subgroup generated by Udef= exp(h ∩W ′). U is certainly an

immersed submanifold of G. We want to show that U is a chart that we can move to anypoint in H.

Well, H and U satisfy the hypotheses of:

Proposition 15.2: Let H be a group, e ∈ U ⊆ H which is a manifold — this is totallygeneral, where the word “manifold” is replaced by almost any geometric category, whereobjects have sheaves of functions sufficiently local — such that the maps µ : U×U → H,i : U −→

u7→u−1H, and Ad(h) : U −→

u7→huh−1H (for h in a generating set of H) have the

following properties:

(a) The preimage of U ⊆ H is open in the domain.

(b) The restriction of the map to this preimage is s/a/h.

Then H has a unique structure as a group manifold such that U is an open submanifold.


The conditions are carried to any composition of maps. So actually Ad(h) is ok for anyh ∈ H.

We begin by further shrinking U . Of course, e ∈ U , so (e, e) ∈ U × U , and so we finde ∈ V ⊆ U such that V 2 ⊆ U . Iterating, we can for any n find a V so that V n ⊆ U . Forus, it’s enough for n = 3, and it will also be convenient to assume that V = V −1.

We view each coset xV as a manifold via Vx·→ xV , with inverse x−1·. So now we

may have an element in multiply cosets, and a priori it might have different manifoldstructures around it. So let W ⊆

openV and consider yW ∩ xV , which corresponds to

x−1yW ∩ V . If that’s empty, it’s certainly an open set. Otherwise, x−1yw = v forsome x ∈ W and v ∈ V , so y−1x = wv−1 ∈ V 2 so y−1xV ⊆ U , and in particular{y−1x} × V ⊆ µ−1(U) ∩ (U × U). So V → y−1xV is continuous and in fact s/a/h. But

45

x−1yW ∩ V is the preimage of W , hence open in V . Thus, the topologies of xV and yVagree on their overlap **indeed, the s/a/h structure**.

So, we put a topology on H by saying that S ⊆open

H if S ∩ xV ⊆open

xV for all x ∈ H.

In particular, each xV receives the subspace topology with respect to this topology, andit’s open. Furthermore, v 7→ y−1xv is s/a/h, and this is the change of coordinates fromthe chart yV to the chart xV . So actually the same functions on the overlap are smoothwith respect to either chart. (We have generalized our notion of “chart”: V is just amanifold, not an open subset of Rn. But if we cover V by honest charts, then we cancover xV , etc.) So H is a “manifold” — well, we didn’t show it’s Hausdorff, but we’llget that for free, knowing that H is a group.

To check that the group structure of H is compatible with this manifold structure, weneed to show that the structure given by right cosets V x gives the same as the left-cosetsdid. Well, all the left cosets are compatible, and all the right cosets are. We need onlycheck that xV and V x are compatible. xV ∩ V x ⊆ xV transports to V ∩ x−1V x, whichby hypothesis is open in V . So xV ∩ V x ⊆

openxV , and all the maps are s/a/h. So the

left and right manifold structures agree.

Finally, we want to check the group structure. (xV )−1 = V −1x−1 = V x−1, and multi-plication is given by µ : xV ×V y → xUy, and we want to show this is s/a/h. But left- orright-multiplication is s/a/h with respect to the left- or right- structure, and V ×V → Uis s/a/h.

The only thing left to show is that it’s Hausdorff. But {e} ⊆ V is closed, and we’re atopological group. The details are an exercise. �

�

Next time, we will review path-connected and simply-connected spaces, and explain the full theoremrelating Lie algebras and Lie groups.


Last time, we saw a subgroup theorem: {h ≤ Lie(G)} ↔ {H ≤ G} where H is connected. Thisis part of a broader association between Lie algebras (finite-dimensional) and Lie groups (simplyconnected).

16.1 Review of algebraic topology

Given a space X with points x, y ∈ X, we define a path from x to y to be a continuous function[0, 1] → X such that 0 7→ x and 1 7→ y. **For want of better notation, I’ll write a path as

46

P : x y.** We can concatenate paths: if x P y

Q z, then we get the concatenation P · Q by

P ·Q(t) =®P (2t), 0 ≤ t ≤ 1/2Q(2t− 1), 1/2 ≤ t ≤ 1

(16.1)

In general, x ∼ y if there’s a path connecting x to y is an equivalence relation. The equivalenceclasses are called path components of X. If X has only one path component, it is called pathconnected. Path connectedness implies connectedness, but not vice versa.

Let A be a distinguished subspace of Y , and f, g : Y → X two functions that agree on A. Thena homotopy f ∼

Ag relative to A is a continuous map h : Y × [0, 1] → X so that h(0, y) = f(y),

h(1, y) = g(y), and h(t, a) = f(a) = g(a) for a ∈ A. For example, a path is a homotopy ofconstant maps **maps from {pt}**. We will only need homotopies of paths, relative to theirendpoints.

Homotopies also concatenate: f ∼Ag is an equivalence relation. Paths are homotopic rel endpoints

to their reparameterizations, and concatenations of homotopies are homotopic. The fundamentalgroupoid of X has as objects the points of X and arrows x→ y the homotopy classes of paths x y.(A groupoid is a category all of whose morphisms have inverses.) Composition is concatenation,and identity is the constant path.

A groupoid is lots of groups. If we pick a point x, the arrows from x→ x form a group π(X,x), andif x and y are path connected, the group at x is isomorphic to the group at y **by whiskering**.If X is path connected, then the groups π(X,x) are isomorphic. Not canonically so: if we pick adifferent path, then we get a different isomorphism, which is by conjugation by a loop.

We say that X is simply connected if π(X,x) is trivial for each x.

Basic examples: S1 is not simply connected, but higher-dimensional spheres Sd are, as is Rn.

A covering space of X is a space E and a “projection” E → X so that locally the covering lookslike a discrete space. I.e. there is a non-empty discrete space S and a covering of X by opens U sothat for each U we have

π−1(U)π

��7777777∼= S × U

��

U

Covering spaces have the

Path-lifting property: Given any path x y and a lift e ∈ π−1(x), then there is a unique pathin E starting at e that projects to x y.

We won’t prove this, but the sketch is that you cover X to locally trivialize E, and lift in each openset, and use compactness of [0, 1].

Homotopy-lifting propert: We can also lift homotopies ∼A

: Y ⇒ X provided Y is locally com-pact.

47

Thus, we get a functor E : π1(X)→ Set which sends x 7→ π−1E (x) and {x y} gives a function by

lifting to each basepoint. In particular, π1(X,x) maps to permutations of π−1E (x).

In particular, if E is simply-connected, then a loop in X is contractible (homotopic to the trivialloop) if and only if it lifts to a loop in E. Noncanonically, we can identify the automorphism groupof E → X with the fundamental group at each point.

Question from the audience: Can you use the deck transformations to make this more canoni-cal? Answer: Maybe, but that’s not actually what I want to talk about. It is an important partof the story, but not of our story.

Assume that X is locally path connected — i.e. that each x ∈ X has arbitrarily small path-connectedneighborhoods — and locally simply connected — that it has a covering by simply-connected opensets — and also that X is path connected. These words are bad, but such is history.

Proposition 16.1: (a) X has a simply-connected covering space X.

(b) X has the universal property: given f : X → Y and a covering π : E → Y , and given achoice of an element from π−1(x) and an element of π−1(f(x)), then

X

π��

f // E

π

��X

f // Y

(c) If X is a manifold, so is X. If f is s/a/h then so is f .

The space X, unique up to isomorphism, is called the universal cover of X.

In particular, a connected manifold satisfies all the conditions.


We sketch the proof. Fix a point x0 ∈ X. Then for each y ∈ X, the homotopy class of pathsx0 y is a point in X. Use locality to construct trivializations, and track consistency onoverlaps.

To check simply-connectedness, it suffices to check the universal property. �

We now turn to the applications of this to Lie groups.

Proposition 16.2: (a) Let G be a connected Lie group, and G its simply-connected cover. Picka point e ∈ G a point over the identity e ∈ G. Then G in its given manifold structure isuniquely a Lie group with identity e such that G→ G is a homomorphism. This inducesan isomorphism of Lie algebras Lie(G) ∼= Lie(G) = g.

(b) G has a universal property. Given any Lie algebra homomorphism α : g → Lie(H), wehave a unique homomorphism φ : G→ H inducing α.

48


(a) If X and Y are simply-connected, then so is X × Y . So by universal property G× G isthe universal cover of G×G. Then lifting the functions µ : G×G→ G and i : G→ Gto G automatically gives the group axioms.

(b) We have α : g → h = Lie(H). This has a graph f ≤ g × h a Lie subalgebra. By thesubgroup theorem, this corresponds to a subgroup F ≤ G ×H (we don’t need simply-connected yet, but if we work in G×H, F might not be the graph of a map). Then

F

f∼→ g !!

⊆ G×H

��G

Use universal property a few times: F ∼= G since it is connected and simply connected,and F is the graph of a homomorphism φ : G→ H.

�

We have almost the full theorem. If g is the Lie algebra of a Lie group G, then we can take itsuniversal cover G. We need:

Theorem 16.3: Ado’s Theorem

Every finite-dimensional Lie algebra g over C is isomorphic to some Lie subgroup of gln(C).

We will prove this later. It’s disappointing, and requires the structure theory of Lie algebras.

Corollary 16.3.1: Every finite-dimensional Lie algebra is g ∼= Lie(G) for a simply-connected G.Then there are functors

{finite-dimensional Lie algebras}Lie(−)

�Grp(−)

{simply-connected Lie groups} (16.2)

And Grp(−) is left-adjoint to Lie(−).


**I was five minutes late.**

17.1 Two-dimensional Lie Algebras

There is an Abelian two-dimensional Lie algebra, with basis X,Y and [X,Y ] = 0. This integratesto three possible groups: R2, R× (R/Z), and (R/Z)2.

49

The other possibility for a two-dimensional Lie algebra, up to a change-of-basis, is [X,Y ] = Y :

−X YadY

adX

We can represent this as X =ñ

1 00 0

ôand Y =

ñ0 10 0

ô, which exponentiates to the group

B =®ñ

a b0 1

ôs.t. a ∈ R+, b ∈ R

´(17.1)

Then B = R+ n R ((b, a)(b′, a′) = (b + ab′, aa′)). B is a simply connected subgroup of GL2(R); itmight be the cover of some other group. Well, if G→ G is a cover, then the kernel is π1(G, e) is adiscrete normal subgroup of G.

Lemma 17.1: A discrete normal subgroup A of a connected Lie group G is in the center: A ≤Z(G). In particular, any discrete normal subgroup is abelian.

And B has no center, by explicit calculation.

In the complex case, there are some subtleties. The simply-connected abelian two-(complex-)dimensional group is C2 under +. But there are lots of ways to mod out. We can mod outby a one-dimensional lattice:

C −→z 7→ez

C× −→qZ

E (17.2)

We pick a q ∈ C× with |q| 6= 1. Then E as a real Lie group is just (R/Z)2, but can have differentholomorphic structures. We have C2, and so we can get different things like C × E, C× × C×,. . . .

In the non-abelian case, we have the same multiplication law

BC =®ñ

a b0 1

ôs.t. a ∈ C×, b ∈ C

´= C× nC (17.3)

This is no longer simply connected. C y C by z · w = ezw, and the simply-connected cover of Bis

BC = Cn C (w, z)(w′, z′) def= (w + ezw′, z + z′) (17.4)

This is an extension:0→ Z→ BC → BC → 0 (17.5)

with the generator of Z being 2πi. Other quotients are BC/nZ.

50

17.2 A dictionary between algebras and groups

Lie Algebra g Lie Group G (with g = Lie(G))

Subgroup h ≤ g Connected Lie subgroup H ≤ G

Homomorphism h→ g H → G provided H simply connected

Module/representation g→ gl(V ) Representation G→ GL(V ) (G simply connected)

Submodule W ≤ V with g : W →W Invariant subspace G : W →W

V g def= {v ∈ V s.t. gv = 0} V G = {v ∈ V s.t. Gv = v}

ad : g y g via ad(x)y = [x, y] Ad : Gy G via Ad(x)y = xyx−1

An ideal a, i.e. [g, a] ≤ a, i.e. sub-g-module A is a normal Lie subgroup, provided G is connectedg/a is a Lie algebra G/A only if A is closed in G

Center Z(g) = gg Z0(G) the identity component of center; this is closed

Derived subalgebra g′def= [g, g] an ideal Should be commutator subgroup, but that’s not closed:

the closure also doesn’t work, although if G is compact,then the commutator subgroup is closed.

Semi-direct product g = h⊕ a with If A and H are closed, then A ∩H is discrete, andh y a and a an ideal G = H n A

Since g-modules are U(g)-modules, we will assume knowledge of algebra modules without muchcomment. We say that a g-module V is simple if there is no submodule W with 0 6= W 6= V . Wesay that g is simple if it is simple as a g module under the adjoint action, i.e. no ideals 0 6= a 6= g.The simply-connected cover G of the exponential of g may have a large center Z; we say that G/Zis “simplest”. It will turn out that the simple Lie groups are all algebraic, and taking them overfinite fields gives most of the simply groups.

If a ≤ g is an ideal and W ≤ V is a submodule, then aW is a submodule. If a, b are ideals, then[a, b] is, e.g. the derived algebra g′ = [g, g]. We can construct the derived series

g ≥ g′ ≥ g′′ ≥ . . . (17.6)

If g is finite-dimensional, this will eventually stabilize. g is solvable if some gn = 0, where we definegi+1 = [gi, gi]. There’s also the lower central series given by gi+1 = [g, gi]:

g ≥ [g, g] = g1 ≥ [g, g1] ≥ . . . (17.7)

Then g is nilpotent if gn = 0 eventually.

51


Last time we had a long list of definitions. Today, we’d like to understand more of the structuretheory of Lie algebras. First:

If V,W are g-modules, there’s a natural way to make V ⊗K W and HomK(V,W ) into g-modules.If we have the Lie group Grp(g) in the background, we can do this via functorial nonsense, but wecan also do it directly algebraically.

Recall that U(g), which whose algebra modules are g-modules, is a bialgebra: we have ∆ : U(g)→U(g)⊗U(g) which on g acts by x 7→ x⊗1 +1⊗x. There’s also the antipode map S : U(g)→ U(g)op

sending x 7→ −x if x ∈ g **the convention seems to be to use an S for the antipode mapin any Hopf algebra; MH uses an “anti-italic” S, and I don’t have wrong-way-slantingfonts, but I do have lots of different script fonts**. We write Aop for the algebra A withthe opposite multiplication; since [−x,−y] = −[y, x], x 7→ −x is an anti-Lie homomorphism, hencean anti-algebra homomorphism on U(g).

With these, we can construct g-representations on V ⊗W and Hom(V,W ). On the former, x(v ⊗w) def= ∆(x)(v ⊗ w) = xv ⊗ w + v ⊗ xw. Moreover, if A y V,W , then A ⊗ Aop y Hom(V,W ) by(x ⊗ y)φ = x ◦ φ ◦ y. So we use the coproduct and antipode to build a map (1 ⊗ S) ◦∆ : U(g) →U(g)⊗ U(g)op. This gives us an action U(g) y Hom(V,W ) by x · φ = x ◦ φ− φ ◦ x.

If g = Lie(G), then the above maps are what we expect. If G y V,W and γ(t) = exp tX,then γ(t)(v ⊗ w) = γ(t)v ⊗ γ(t)w, and taking d

dt

∣∣∣t=0

gives Xv ⊗ w + v ⊗ Xw. Differentiatingγ(t) · φ = φ(t)φγ(−t) gives the other map.

Moreover, we have a meta-theorem: there are various functorial constructions, e.g. V ⊗ W ∼=W ⊗ V and Hom(U ⊗ V,W ) ∼= Hom(U,Hom(V,W )), and W : V 7→ Hom(Hom(V,W ),W ). Thenthese are all g-module homomorphisms with the above representations. To prove this requiresshrinking the infinite list of functors to a finite list that generates them **using various coherenceresults**.

We can in particular construct g-invariant maps:

Hom(V,W )g def= {φ : V →W : x ◦ φ = φ ◦ x ∀x ∈ g} = Homg(V,W ) (18.1)

18.1 More structure theory

Recall that for g finite-dimensional we defined three series **should call these “sequences”**:

• the derived series g0 def= g and gn+1 def= [gn, gn]

• the upper central series g0def= g and gn+1

def= [g, gn]

Then we say that g is solvable if some gn = 0, nilpotent if some gn = 0, and semisimple if 0 is theonly solvable ideal. It’s equivalent to replace the word “solvable” in the definition of “semisimple”

52

with the word “abelian”, since if r 6= 0 is an ideal, then rn are ideals, and the last non-zero rn isabelian.

Each gi+1 is normal in the previous gi; then gi/gi+1 is abelian. But we don’t want to think ofthese as ideals to much. Any K-subspace between gi and gi+1 are all ideals, because conjugatingany gi ≥ h ≥ gi+1 by gi lands in gi+1 and in particular in h. So we can interpolate between gi andgi+1 by one-dimensional extensions. I.e. if gi is solvable, then there is a sequence ai of subalgebrasso that ai/ai+1 is one-dimensional. This feels like a solvable group. The (simply-connected) groupversion in the simply connected case should be that we can find a sequence of closed subgroups,each with codimension 1, so the quotients are all abelian.

There is a similar story for nilpotence, which is easier to analyze:

Proposition 18.1: If g is nilpotent then it is solvable.


gi ≤ gi �

Proposition 18.2: If g 6= 0 is nilpotent, then Z(g) 6= 0.


The last non-zero gl is central. �

Proposition 18.3: Any subalgebra or homomorphic image of a solvable (nilpotent) g is solvable(nilpotent). Moreover, if a is an ideal in g and if a and g/a are solvable, then so is g: extensionsof solvable Lie algebras are solvable. The corresponding statement about nilpotence is notquite true, but if a is nilpotent and g y a nilpotently, then g is nilpotent. In particular, if a

is central, this is true: a central extension of a nilpotent Lie algebra is nilpotent.


The derived and central series of subquotients are subquotients of the derived and centralseries. For the second statement, we start taking the derived series of g, eventually landingin a (since g/a→ 0), which is solvable. The nilpotent claim is similar. �

Example: Let g = 〈X,Y : [X,Y ] = Y 〉 be the two-dimensional nonabelian Lie algebra. Theng1 = 〈Y 〉 and g2 = 0, but [g, 〈Y 〉] = 〈Y 〉 so g2 = g1.

There is also the lower central series **with even worse indexing**: 0 ≤ Z(g) ≤ z2 ≤ . . . givenby z0 = 0 and zk+1 = {x ∈ g : [g, x] ⊆ zk}.

Proposition 18.4: Some zn = g if and only if g is nilpotent.

Theorem 18.5: Engel’s Theorem (wrong)

If g ≤ gl(V ) is nilpotent for V 6= 0 finite-dimensional, then there exists a non-zero invariantvector.

Corollary 18.5.1: Then there is a basis of g so that every matrix in g is upper-triangular.

53

Proof of Engel’s Theorem (wrong):

We can assume that g is simple by induction on dimension: if not, then it has a minimal non-zero submodule, and g acts nilpotently on the submodule, to a quotient of g kills a non-zerovector in the submodule, hence g does. Moreover, we can assume that g 6= 0, otherwise thetheorem is trivial, and since g is nilpotent, Z(g) 6= 0.

By the way, we will end with a contradiction to a bunch of assumptions that we can makewithout loss of generality. It is a weird proof.

Anyway, pick X ∈ Z(g). We should make part of the induction that X is nilpotent in itsaction on V (we will either justify this later or throw out the whole proof and fix it nexttime). Since X is singular, kerX is non-zero, and a submodule since X is central. HencekerX = V , but we assumed without loss of generality that g acts faithfully. �

Question from the audience: Certainly just being central does not imply nilpotence. Answer:No. Let’s throw out the whole proof, and look at corollaries. Question from the audience:For example, g is the one-dimensional subalgebra of gl generated by the identity. This is certainlynilpotent as a Lie algebra, and does not kill anything in V . Answer: Ah, so the proof is right andthe theorem is wrong.

Theorem 18.6: Engel’s Theorem (fixed)

If g ≤ gl(V ) acts by nilpotent matrices, then there is a vector v 6= 0 in V such that gv = 0.

Question from the audience: Then why is Z(g) 6= 0? Answer: Well, the proof is stillwrong.

Question from the audience: Were you going to show that a nilpotent Lie algebra has a nilpotentrepresentation? Answer: Yes, the adjoint action. Indeed,

Remark: If g is a nilpotent Lie algebra, say gn = 0, then for any sequence X1, . . . , Xn ∈ g, wehave (adX1) . . . (adXn) = 0. In particular, adX is nilpotent for any X ∈ g. Engel’s theoremimplies the converse. A Lie algebra in which every element acts nilpotently on a faithfulrepresentation is necessarily nilpotent.

Corollary 18.6.1: If g acts nilpotently on V , then there is a basis of V in which every X ∈ g isstrictly upper triangular.


Pick a killed vector, and use that as one basis element; proceed by induction. �

54


19.1 Engel’s Theorem and Corollaries

We begin by fixing Engel’s Theorem.

Theorem 19.1: Engel’s Theorem (fixed)

If g is a finite-dimensional Lie algebra acting on V by nilpotent endomorphisms, and V 6= 0,then there exists a non-zero vector v ∈ V such that gv = 0.

This is even true for V infinite-dimensional. Remembering this gets the right induction.


Lemma: If g y V,W by nilpotents, then g y V ⊗W and g y Hom(V,W ) by nilpotents.

It suffices to look at the image of g in gl(V ) = Hom(V, V ). **MH writes gl(V ) ≤ Hom(V, V ),but isn’t it everything?** Then ad : g y g is by nilpotents.

Let’s pick v 6= 0 in V such that dim(h) is maximal, where h = {x ∈ g : xv = 0}. Then h ≤ g

is a subalgebra, and we want to show that h = g, so we supposed the contrary h � g. Byinduction on dimension, the theorem holds for h. Well, h y g/h by nilpotents, so we findx ∈ gr h **so its image in g/h is non-zero** where [h, x] ≤ h **so the image of h · xis zero**. Then h1 = 〈x〉+ h is a subalgebra of g.

Let U = {u ∈ V : hu = 0} 6= 0 **because it kills the v in the previous paragraph**.Then U is an h1-submodule: hxu = [h, x]u+ xhu = 0u+ x0. Since x is nilpotent, x|U kills avector v 6= 0 in U . So h1v = 0, contradicting maximality of h. �

Corollary 19.1.1: If g y V by nilpotents and V is finite dimension, then there’s a basis ov V inwhich g is strictly upper triangular.

Corollary 19.1.2: If adX is nilpotent forall x ∈ g finite-dimensional, then g is a nilpotent Liealgebra.

Proposition 19.2: Let V be a simple g-module (also called “irreducible” — no proper submodules— and V 6= 0). If an ideal a ≤ g acts nilpotently on V then a kille V .


a kills a vector, but a is an ideal, so space of vectors it kills is a non-zero g-submodule. �

Now let V be finite dimensional with a g action. We find a Jordan-Holder series

0 < M1 < M2 < · · · < Mn = V (19.1)

where each Mi is a g-submodule, and Mi+1/Mi is simple. Then

Corollary 19.2.1: An ideal a ≤ g acts by nilpotents on V if and only if it kills all Mi+1/Mi.

55

In particular, there is a unique largest such ideal. Be careful: this is not the set of all elements ofg that act nilpotently, just the largest ideal all of whose elements act nilpotently.

Proposition 19.3: Any nilpotent ideal a ≤ g acts nilpotently on g.


[a, g] ≤ a. Define a0 = a and ai+1 = [a, ai]. Eventually an = 0. �

Corollary 19.3.1: g has a largest nilpoent ideal: the nilpotency ideal of ad.

We will say soon that g has a largest solvable ideal, using the fact that extensions of solvables aresolvable. But extensions of nilpotents are not necessarily nilpotent, so we should remember theabove corollary.

If g y V finite-dimensional, then we get a symmetric bilinear for bV (in the book called “B”, butwe want to save that letter for the Baker-Campbell-Hausdorff formula) on g, by βV (x, y) = trV (xy).(Symmetric is by cyclicity.)

We define the radical of βV to be radβV = {x s.t. βV (x, g) = 0}. **I would call this the“kernel”.**

Proposition 19.4: If a ≤ g acts nilpotently on V , then a ≤ radβV .


Consider a Jordan-Holder series again. �

19.2 Solvability

Everything above works over any characteristic. We will sometimes need characteristic 0 and/oralgebraic closure for understanding solvability. But solvability is the natural condition when wethink of Lie algebras as coding for Lie groups.

Proposition 19.5: Every finite-dimensional g has a largest solvable ideal, called rad g.

This is maybe an unfortunate conflict of notation with the radical of a bilinear form, but they arerelated.


If ideals a, b ≤ g are solvable, then a + b is solvable, since we have an exact sequence ofg-modules

0→ a→ a + b→ (a + b)/a→ 0 (19.2)

which is also an extension of a solvable algebra by a solvable ideal. �

Theorem 19.6: Lie’s Theorem

Let g be solvable and g y V 6= 0. Assume that the ground field K contains eigenvalues ofthe actions of all x ∈ g. Then V has a one-dimensional submodule.

56


Without loss of generality g 6= 0; then g′ 6= g by solvability. Pick any g ≥ h ≥ g′ a codimension1 subalgebra, and since h ≥ g′ it is an ideal. Pick x ∈ gr h, so g = 〈x〉+ h.

As a subalgebra, h is also solvable, and by induction on dimension h y V has a one-dimensional submodule 〈~e〉: if h ∈ h, then h · ~e = λ(h) e for some linear λ : h→ K.

Let E = K[x]~e for x ∈ g r h as above. Then E = U(g)~e, and each 〈1, x, . . . , xm〉~e is anh-submodule. Why? We induct on m:

hxm~e = xmh~e︸︷︷︸=λ(h)xm~e

+∑

k+l=m−1

xk[h, x]xl~e︸︷︷︸∈〈1,...,xm−1〉~e by induction

(19.3)

This also shows that E is a generalized eigenspace with eigenvalue λ(h) for all h ∈ h, and sotrE h = (dimE)λ(h), by working in a basis where h is upper triangular. But 0 = tre[h, x] soλ([h, x]) = 0, and so (19.3) shows by induction on m that E is an actual eigenspace.

Question from the audience: You need to make an assumption on characteristic to divideby dimE? Answer: Yes, for this proof. Possibly not for the theorem, but we will needassumptions on characteristic later, so we might as well include it here.

Now take 〈~v〉 where ~v ∈ E is an eigenspace of X, and we’re done. �

Corollary 19.6.1: If g is solvable, g y V finite dimensional, and K has characteristic = 0 and alleigenvalues are in K, then V has a basis in which g is upper diagonal.

This should be our model: if g is upper-triangular, then g′ is strictly upper triangular, and brack-eting moves us further away from the diagonal, so upper-triangular Lie algebras are solvable in anobvious way.

Corollary 19.6.2: If g is solvable, and K with characteristic 0 is algebraically closed, then everysimple finite-dimensional g-module is one-dimensional.

So solvable algebras have the simplest and most complicated modules: every module is an extensionof one-dimenional ones, but extensions are hard. The category of g-modules is very non-semi-simple.If g is semi-simple, we’ll see that every module is a direct sum of simples. They are two sides of thestory: every Lie algebra is a combination of a solvable and a semi-simple, so we need to understandboth representation theories.


We continue our discussion of solvable Lie algbras. Various things we will say require a characteristic-0 hypothesis. We will have some counterexamples in the homework.

We will speak more about corollaries of Lie’s theorem, and then discuss the Killing form.

57

20.1 Solvable algebras (char = 0)

Lie’s theorem has various corollaries. Last time we say that if g is solvable, then it acts upper-triangularly.

Corollary to Lie’s Theorem: If g is solvable, then g′ acts nilpotently on any finite-dimensionalV .

Remark: We don’t need K = K for this corollary, even though we did need it for Lie’s theorem.Because if we have g/K and K ≤ L a field extension, then L ⊗K g has many of the sameproperties as g: e.g. it has the correctly extended series, and if g y V nilpotently, thenL⊗K g y L⊗K V nilpotently (by Engel’s Theorem).

Corollary to Lie’s Theorem: Engel’s theorem says that on any finite-dimensional module hasa largest g has a largest ideal acting nilpotently. We can pin this down further in the solvablecase: if g is solvable, then its ad-nilpotent elements include all of g′, and hence form an ideal.

20.2 Killing Form

Recall, for any finite-dimensional g-module V we get a symmetric bilinear form βV (x, y) def= trV (xy).This form is invariant, in the sense that

− (βV ([z, x], y) + βV (x, [z, y])) = 0 (20.1)

The minus-sign is because we’re really acting on Hom, but it’s 0, so we can drop it. What’s goingon is that we have maps that are gl(V )-invariant:

End(V )⊗ End(V ) ◦→ End(V ) tr→ K (20.2)

And also g→ End(V ) is g-invariant.

We define the Killing form to be β = β(g,ad) the trace form on the adjoint representation g yg.

Remark: f W ≤ V is a g-submodule, then βV = βW + βV/W .

In particular, if a ≤ g is an ideal, then β(g/a,ad)|a×g = 0. So β|a×g = βa|a×g, hence the Killing formof a is β|a×a.

We define the radical (or “kernel”) to be radβV = {x : βV (x,−) = 0}.

Proposition 20.1: radβV is an ideal


βV gives a map g → g∗, and if βV is invariant, then this is a g-module homomorphism, andthe kernel is radβV , hence a submodule and thus an ideal. �

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Proposition 20.2: Corollary to Engel’s Theorem

If a ≤ g is an ideal and a acts nilpotently on V , then a ≤ radβV .

Question from the audience: Why? Answer: Look at the Jordan form of the matrices: theyare in block upper triangular form. But if a acts nilpotently, then it is strictly-upper-triangular,and the product of a strictly-upper-triangular and an upper-triangular is strictly upper-triangular,hence traceless.

In particular, without using Lie’s theorem and the characteristic worries:

Corollary 20.2.1: If β is nondegenerate (i.e. radβ = 0), then g is semisimple. (Otherwise, anabelian ideal is in the radical.)

Let’s dispense right now by the problem of algebraic closure. We saw that the extension L ⊗K g

has the same series. Moreover, radβg = 0 iff radβg⊗L = 0.

20.3 Jordan Form

Proposition 20.3: Let V be finite-dimensional over algebraically closed K. Then

(a) Every x ∈ gl(V ) has a unique Jordan decomposition x = s+n, where s is diagonalizable,n is nilpotent, and they commute (or equivalently one must commute with x).

(b) s, n ∈ xK[x] — polynomials in x with no constant term.


Existence in (a) is implied by Jordan form. It’s not so obviously unique. Looking at (b), letx be a Jordan block:

x =

λ 1

λ. . .. . . 1

λ

Then

K[x] =

a1 a2 . . . an

a1. . .

.... . . a2

a1

If λ 6= 0, then xK[x] = K[x], and if λ = 0, then s, n ∈ xK[x]. We remark that restricting to aJordan block is a ring homomorphism.

Let λ1, . . . , λk be the eigenvalues of x, and find i 7→ Pi(x) ∈ K[x] such that for all j, Pi(x)−δijhas λj as a root of multiplicity >> 0. Then Pi(Y ) = δijI for Y a Jordan block with eigenvalue

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λj , and ∑i

Pi(X)Qi(X) ∈ xK[x]

**I’m not following, just copying the board. Please take with a grain of salt thisproof.**

For uniqueness, let x = s′ + n′. Then s′ + n′ = s + n, and so n′ − n = s′ − s, but n′ − n isnilpotent, and s′ − s is diagonalizable, but only 0 is nilpotent and diagonalizable.

Question from the audience: Why do s and s′ commute? Answer: Because s′ commuteswith x, and s is a polynomial in x. �

We now move to an entirely unmotivated piece of linear algebra:

Lemma 20.4: Let V be finite-dimensional over algebraically closed K of characteristic 0. LetB ⊆ A ⊆ gl(V ) be any subspaces, and define T = {x ∈ gl(V ) : [x,A] ⊆ B}. Then if t ∈ Tsatisfies trv(tu) = 0 ∀u ∈ T , then t is nilpotent.


Let t = s+n. We want to show that s = 0. Fix a basis {ei} in which s is diagonal: sei = λiei.Then let {Eij} be the basis of matrix units for gl(V ). Then (ad s)Eij = (λi − λj)Eij .

Now let Λ = Q{λi} be the finite-dimensional Q-vector space in K. We consider an arbitraryQ-linear functional f : Λ→ Q, and we want to show it’s 0.

Well, f(λi)− f(λj) = f(λi − λj), and chose a polynomial P (x) ∈ K[x] so that P (λi − λj) =f(λi)− f(λj). (So P (0) = 0 and P ∈ K[x].)

Now we define u ∈ gl(V ) by uei) = f(λi)ei, and then (adu)Eij = (f(λi) − f(λj))Eij =P (ad s)Eij . So adu = P (ad s).

But what have we done? ad t = ad s+adn, and ad s, adn commute, and ad s is diagonalizableand adn is nilpotent. So this is the Jordan form, and we have ad s = Q(ad t) for somepolynomial Q. Then adu = P ◦Q(ad t), and since every power of t takes A into B, we have(adu)A ⊆ B, so u ∈ T .

But then tu is upper-triangular, and the diagonal parts multiply, so 0 = tr(tu) =∑λif(λi).

But we apply f to this: 0 =∑

(f(λi))2 ∈ Q, so f(λi) = 0 for each i. Thus f = 0. �

Theorem 20.5: (Cartan)

Let V be finite-dimensional over characteristic 0 field K. Then g ≤ gl(V ) is solvable iffβV (g, g′) = 0, i.e. g′ ⊆ radβV

The proof is next time, and we can extend by scalars.

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We’d like to take up today two Cartan criteria, for solvability and nilpotency of Lie algebras. Apartfrom having technical uses of these, they’re also basic tools.

21.1 Cartan’s Criteria

We recall the lemma from last time:

Lemma 21.1: Let V be finite-dimensional over K algebraically closed of characteristic zero. Givensubspaces B ⊆ A ⊆ gl(V ) arbitrary subspaces, let T = {x ∈ gl(V ) s.t. [x, a] ≤ B}, clearlya Lie subalgebra. Then if t ∈ T and trV (tu) = 0 for every u ∈ T , then t is nilpotent. I.e.radβV |T×T consists of nilpotents.

Theorem 21.2: (Cartan)

Let V be finite-dimensional over K of characteristic 0. Then g ≤ gl(V ) is solvable iffβV (g, g′) = 0 (i.e. g′ ≤ radβV ).


We can extend scalars and assume that K is algebraically closed, thus we can use the lemma.

The forward direction follows by Lie’s theorem: we can find a basis of V in which g acts byupper-triangular matrices, and hence g′ acts by strictly upper-triangular matrices.

For the reverse, we’ll show that g′ acts nilpotently, and hence is nilpotent by Engel’s theorem.We use the lemma, taking V = V , A = g, and B = g′. Then T = {t ∈ gl(V ) s.t. [t, g] ≤ g′},and in particular g ≤ T , and so g′ ≤ T .

Question from the audience: What’s V ? Answer: V is the vector space we’re startingwith. Question from the audience: Oh, yeah, that was a dumb question. Answer: Noit wasn’t. It confused me until I figured out what it was.

So if [x, y] = t ∈ g′, then trV (tu) = trV ([x, y]u) = trV (y[x, u]) by invariance, and y ∈ g and[x, u] ∈ g′ so trV (y[x, u]) = 0. Hence t is nilpotent. �

We have corollaries, all of which are only in characteristic 0:

Corollary 21.2.1: g is solvable iff g′ ≤ radβ.


Take V = g with the adjoint representation. g may not be a subalgebra of gl(V ), but takeg = g/Z(g) (mod out by the center), and g ↪→ gl(V ). Of course, g is a central extension ofg, and so g is solvable iff g is, and by theorem says that g is solvable iff βg(g, g) = 0. But βg

factors through βg:

βg = {g× g/Z(g)−→ g× g

βg→ K} �

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Corollary 21.2.2: radβ is solvable, i.e. radβ ≤ rad g (in characteristic 0 and g finite-dimensional)

Corollary 21.2.3: (Cartan)

g is semisimple iff radβ = 0.


The reverse direction is true in any characteristic. The forward direction follows from theprevious corollary. �

Corollary 21.2.4: g is semisimple iff any extension by scalars is.

Note: g/ rad g is semisimple.

So solvable and semisimple algebras are the building blocks of everything. We will later see

Levi’s Theorem: Let r = rad g. Then g = s⊕ r where s is a semisimple subalgebra of g. This isonly in characteristic 0.

21.2 Three-dimensional Lie algebras.

We already classified the two-dimensional algebras. Some three-dimensional algebras are extensionsof smaller things. We have a few more:

• Heisenberg: Z is central and [X,Y ] = Z. This is nilpotent.

• sl2: Our (ordered) basis is E,H,F , and the relations are [H,E] = 2E, [H,F ] = −2F , and[E,F ] = H. Let’s understand the Killing form of sl2.

adH =

20−2

adE =

0 −2 00 0 10 0 0

adF =

0 0 0−1 0 00 2 0

Then we can compute β, which is symmetric **and not really a matrix, but actually a2-tensor, in case we care about the variance**:

βsl2 =

0 0 40 8 04 0 0

(21.1)

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This is non-singular, so sl2 is semisimple. In fact, it’s simple. This is different over charac-teristic 2. Question from the audience: I can change the presentation to get rid of the 2s**h = H/2, e = E/

√2, f = F/

√2**. Maybe over characteristic 2 this doesn’t act on K2?

Answer: You still get 2s in β. I don’t think you can win.

Proposition 21.3: In characteristic 0, if g is semisimple than every ideal a and quotient g/a issemisimple.

By the way, the converse of this, that extensions of semisimples are semisimple, is true in anycharacteristic.


We know that β is nondegenerate. Let a⊥ be the orthogonal subspace to a with respet to β.Then a⊥ = ker{x 7→ β(−, x) : g → Hom(a, g)}, so a⊥ is an ideal. Then a ∩ a⊥ = radβ|a ≤rad a, and hence it’s solvable and hence is 0. So a is semisimple, and also a⊥ is. In particular,the projection a⊥

∼→ g/a is an isomorphism of Lie algebras, so g/a is semisimple. �

Notice that [a, a⊥] ≤ a ∩ a⊥ = 0. We iterate, e.g. by taking a to be a minimimal and hence simpleideal. Hence:

Corollary 21.3.1: Every semisimple g is a direct product g = g1 × · · · × gm of simple nonabelianalgebras.

In particular, sl2 must be simple.

21.3 Casimir operator

The representation theory of solvable and semisimple algebras are very different. If g is semisimple,then every finite-dimensional module is a direct sum of irreducible ones. If g is solvable, then everyfinite-dimensional module is an extension by one-dimensional modules.

In any case, to get a handle on the representation theory, we introduce the Casimir operator.

We begin with some linear algebra:

Proposition 21.4: Let 〈, 〉 be a nondegenerate not-necessarily-symmetric bilinear form on finite-dimensional V . Let (xi) and (yi) be dual bases, so 〈xi, yj〉 = δij . Then θ =

∑xi⊗ yi ∈ V ⊗V

depends only on the form 〈, 〉. If z ∈ gl(V ) leaves 〈, 〉 invariant, then θ is also invariant.

**I should go through and fix the indices.**


Let {ξi} be the dual basis of V ∗ dual to xi: ξixj = δij . Then 〈, 〉 induces ψ : V → V ∗ viav 7→ 〈−, v〉, and ψ(yi) = xi. In particular, θ =

∑xi ⊗ yi = (idV ⊗ ψ−1)(

∑xi ⊗ ξi). So we

have only to show that∑xi⊗ ξi does not depend on the basis, but λ =

∑xi⊗ ξi ∈ V ⊗V ∗ ∼=

Hom(V, V ), and indeed λ = idV .

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Now, End(V ) ∼= V ⊗ V ∗ id⊗ψ−1

−→ V ⊗ V takes idV 7→ θ. If z leaves 〈, 〉 invariant, it commuteswith ψ, and hence with id⊗ ψ−1, so θ is invariant. �

Let β be a nondegenerate invariant (symmetric) form on finite-dimensional Lie algebra g. Pickdual basis {xi} and {yi} in g. Then θ

def=∑xi ⊗ yi ∈ g ⊗ g

mult−→ U(g) and we define the Casimiroperator of β to be cβ =

∑xiyi the image of θ in U(g). But if β is g-invariant, then cβ is central

in U(g).

If β = βV **do we need to assume g y V faithfully so that β is nondegenerate? Thisis necessary but not sufficient, e.g. if g is the strictly-upper-triangular things. Sowe must have some sort of semisimplicity assumption on g?**, then cβ acts on V , andtrV (cβ) = tr(

∑xiyi) = dim g, which is non-zero if characteristic is 0. So the casimir operator

distinguishes V from the trivial representation.

**Here this is in indices: Let βij be nondegenerate bilinear form on V finite-dimensional.Then θij the inverse form (defined by θikβkj = δij and βikθ

kj = δji ) exists in V ⊗ V . Ifzimz

jnβij = βmn, then zmi z

nj θ

ij = θmn.

We define cβ to be the image of θij in U(g) when βij is an invariant form on g. Herei, j are indices on g, which is the vector space V in the previous paragraph. Let Γijkbe the structure tensor [g, h]i = Γijkg

jhk. A representation ρ : g y V is a tensor ρiνµ,

where µ, ν are V -indices, such that ρiνµρj

µλ − ρjνµρi

µλ = Γkijρk

νλ. We can extend ρ to U(g)

by extending the g index to a U(g) (multi-)index. Let βij be the trace form of therepresentation ρi

νµρj

µν , and c

~k its casimir. Then ρ~kνµc~k is central — ρ~k

νµc~kφµλ = ρ~k

µλc~kφνµ —

and ρ~kµµc~k = dim g = δii for i a g-index. On the other hand, c∅ = 0, i.e. c has no terms in

degree 0 in U(g), so c acts as 0 in the trivial representation τi = 0.**


Today we launch forward towards the “complete reducibility” theorem of modules over certain Liealgebras. We will approach this from a sideways direction, speaking in the language of homologicalalgebra. It’s possible to avoid the words “homological algebra”, but the classical proofs in hindsightare homological, with the useful words erased.

First we recall from last time: If g y V finite-dimensional, we get a trace form βV (x, y) def= trV (xy),and we suppose this is nondegenerate. Of course, βV is invariant: βV ([x, y], z) + βV (y, [x, z]) = 0.Forshadow: a semisimple algebra is the direct sum of simples. If g were simple, then V is eitherthe trivial module or radβV is an ideal that’s not everything, hence 0. So semisimple algebrashave plenty of nondegenerate forms. In any case, we define the Casimir operator cβV =

∑xiyi ∈

U(g) where {xi} and {yi} are dual bases of g for βV : βV (xi, yj) = δij . This has some niceproperties:

1. cβV only depends on βV .

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2. cβV ∈ Z(U(g))

3. cβV ∈ U(g)g, i.e. it acts as 0 in K.

4. trV (cβV ) =∑

trV (xiyi) = dim g.

22.1 Review of Ext:

Let U be an associative algebra, M,N (left-)modules. A free module F is a possible-infinite directsum of copies of U y U . A free resolution of M is a complex F• **MH writes F •, but thenuses lower indices** of free module

· · · → Fkdk→ Fk−1 → · · · → F1 → F0 → 0

which is exact everywhere except at 0, where its homology if M . I.e. the augmented complexF• →M → 0 is exact. You can always build such a thing: take generators of M and use those forF0, take generators of the kernel F0 →M and use those for F1, etc.

We recall that a complex is a sequence above so that di ◦ di+1 = 0, and given a complex we canbuild homology modules Hi = ker di/ im di+1.

We remark that we are using homology indexing, which is to say that our sequence goes to the leftand we use lower indices. We could call the Fi instead F−i, and then we’d call Hi H

−i and say“cohomology”.

Let’s apply the functor HomU (−, N), which is contravariant, to F •:

HomU (F •, N) = Hom(F0, N) δ1→ Hom(F1, N) δ2→ . . . (22.1)

We define ExtiU (M,N) = H i(HomU (F•, N)). In particular, Ext0U (M,N) = Hom(M,N).

It’s clear that for each choice of a free-resolution of M , we get a functor Ext•(M,−). How aboutin M? Let M → M ′ be a U -morphism, and F ′• a free resolution of M ′. Then by freeness we canextend the morphism M →M ′ to a chain morphism, unique up to chain homotopy:

. . . // F1

��

// F0//

�� M

��. . . // F ′1

// F ′0// M ′

(22.2)

Since chain homotopies induce isomorphisms on Hom, this really is functorial, and in particularif M = M ′ and we take different free resolutions, then the above shows that Ext•(M,N) is well-defined.

There are fancier versions of all this that say you can take projective resolutions of M or injectiveresolutions of N , but for those you should read a book on homological algebra: we won’t needthem.

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Let 0 → A → B → C → 0 be short-exact. Then Homing preserves left-exactness but can killright-exactness. But the following long sequence is exact:

Ext0(A,N)GF@AEDBC

pp````

Ext0(B,N)oo Ext0(C,N)oo 0oo

Ext1(A,N)GF@AEDBC

pp````

Ext1(B,N)oo Ext1(C,N)oo

. . . Ext2(A,N)oo Ext2(B,N)oo Ext2(C,N)oo

(22.3)

The existence of the connecting functors is special.

We are working towards a theorem of Weyl, called complete reducibility: If g is semisimple overK of characteristic 0, then every finite-dimensional g-module is a direct sum of irreducible ones.A module is irreducible if it has no proper submodule (compare to indecomposible, not the directsum of submodules, for which the statement is obvious). Weyl’s theorem is not true for solvablemodules.

Let’s understand this theorem in Ext language. If 0 → A → B → C → 0 is an extension withA irreducible, then let’s use the long-exact sequence (22.3) with N = A. Then we can takeHom(A,A) → Ext1(C,A), and map idA 7→ α, called the characteristic class of the extension.If α = 0, then idA ∈ im(Hom(B,A) → Hom(A,A)), and so the sequence splits and cosplits:B ∼= A⊕ C.

It turns out that Ext1(C,A) bijectively classifies (equivalence classes) of extensions. But in partic-ular, if Ext1(M,N) = 0 for every finite-dimensional representations M and N , then every finite-dimensional representation is complete reducible.

Lemma 22.1: If F is a free U(g)-module, then so is F ⊗K N for any N .

Already we’re using something special about U(g), that it is a bialgebra. Otherwise, the tensorproduct of modules is not a module.


F =⊕U(g), and tensor distributes over (even infinite) direct sums, so we reduce to F = U(g).

How does U(g) y G = U(g) ⊗ N? Let X ∈ g and u ⊗ n ∈ U(g) ⊗ N . Then X(u ⊗ n) =Xu⊗ n+ u⊗Xn, where Xu is the product in U(g), and Xn is the action g y N .

We put a filtration on G by G≤n = U(g)≤n ⊗K N . This makes G into a filtered module:

U(g)≤kG≤l ⊆ G≤k+l (22.4)

I.e. grG is a grU(g)-module, but grU(g) = S(g), and S(g) acts through the first term, soS(g)⊗N is a free S(g)-module, by picking any basis of N .

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So, let {nβ} be a basis of N and {Xα} a basis of g. Then {X~αnβ is a basis of grG = S(g)⊗N ,hence also a basis of U(g)⊗N . Thus, by using the PBW theorem a couple times implicitly,we see that U(g)⊗N is free. �

The philosophy here is that the homology of U(g) should be a slight perturbation of the homologyof S(g), which is easy to control.

Corollary 22.1.1: IfM andN are finite-dimensional g-modules, then Exti(M,N) ∼= Exti(K,Hom(M,N)) ∼=Exti(Hom(N,M),K).

**MH writes M∗⊗N for Hom(M,N), but said “Hom”, using finite-dimensionality.**


It suffices to prove the first

Let F• →M be a free resolution. Well, a U(g)-module homomorphism is exactly a g-invariantlinear map:

HomU(g)(F,N)• = HomK(F•, N)g (22.5)

= HomK(F• ⊗K N∗,K)g (22.6)

= Ext•(M ⊗N∗,K) (22.7)

Using the finite-dimensionality of N and the lemma that F • ⊗ N∗ is a free resolution ofM ⊗N∗. �

Lemma 22.2: If M,N are f.d. U -modules and c ∈ Z(U) such that the characteristic polynomialsof c on M and N are relatively prime, then Exti(M,N) = 0 for all i.

Roughly, use functoriality: if c is central, then it acts on Exti and must satisfy its characteristicpolynomials from both M and N , hence must satisfy 1, which is a linear combination of thecharacteristic polynomials. I.e. 1 acts on Ext by 0, so Ext = 0.


**I was a few minutes late, copying the first few results from the board, and filling inthe first few proofs (until the main theorem) from Dustin.**

Lemma 23.1: If M and N are f.d. U -modules and c ∈ Z(U) s.t. the characteristic polynomials fand g of cyM,N are coprimes, then Exti(M,N) = 0 ∀i.


From last time: f(c) kills M hence Exti(M,N), and g(c) kills N and hence Exti(M,N), and1 = af + bg hence kills Exti(M,N). �

Philosophy: think of Z(U) as a polynomial ring, and then we’re saying that M and N have disjointsupport.

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Theorem 23.2: Schur’s Lemma

Let N by a simple non-zero U -module and α : N → N a U -homomorphism; then α = 0 or αis an isomorphism.


imα is a submodule, hence either 0 or the whole thing. If im = 0, then we’re done. Ifimα = N , then kerα 6= 0, so kerα = N by simplicity, and α is an isom. �

Corollary 23.2.1: LetM,N f.d. simple U -modules, c ∈ Z(U) killsM but notN , then Exti(M,N) =0 ∀i.


By Schur, c acts invertible on N , so all its eigenvalues over the algebraic closure are non-zero.But eigenvalues of c on M are 0, so the characteristic polynomials are relatively prime. �

Theorem 23.3: Let K be characteristic 0 and g semisimple. Then Ext1(M,N) = 0 for every M ,N f.d. g-modules.


Reduce to the case M = K and N simple. If N¬(K, trivial), then g = g1 × · · · × gk actsnon-trivially on N . So some gi acts non-trivially, hence βN does not vanish on gi by Cartan’scriterion. So radgi βN = 0, giving a casimir c ∈ Z(U(gi)) ⊆ Z(U(g)), and trN (c) = dim gi 6= 0but c kills K. So if N is non-trivial, we’re done by the corollary.

There are multiply ways to complete the proof, which requires calculating Ext1(K,K) (this isthe N = K case). In one way, we use the fact that Ext1(K,K) classifies extensions 0→ K→

L → K → 0, and hence L is two-dimensional and g must act likeñ

0 ∗0 0

ôon L, but then g

acts nilpotently and g is semisimple, so g kills L and the only extensions is the trivial one.

The other way to do this is to use an exact sequence 0→ I → U(g)→ K→ 0. We won’t dothis, because we will calculate the entire resolution of K over U(g). �

Corollary 23.3.1: (Weyl)

Every finite-dimensional representation of a semisimple g over characteristic 0 is completelyreducible.

Corollary 23.3.2: (Whitehead)

If g is semisimple over char zero, then Exti(M,N) vanishes for all i if M and N are finite-dimensional non-isomorphic simples.

So this tells us almost all the higher Exts, just not those of K with K, etc. So we explore how tocompute those:

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23.1 Computing Exti(K, M)

We will compute a free-resultion of K. We will use Levy’s theorem: if g is f.d. over char zero, theng/ rad is semisimple, and in fact g is a direct product of its nilpotent and semisimple parts.

Proposition 23.4: K has a free U(g)-resultion given by

· · · → U(g)⊗∧2g→ U(g)⊗ g→ U(g)→ K→ 0 (23.1)

where all ⊗ are over K and dk : U(g)⊗∧kg→ U(g)⊗∧k−1g is given by

dk(x1 ∧ · · · ∧ xk) =∑i

(−1)i−1xi ⊗ (x1 ∧ . . . xi · · · ∧ xk)

−∑i<j

(−1)i−j+11⊗ ([xi, xj ] ∧ x1 ∧ . . . xi . . . xj · · · ∧ xk) (23.2)

Some examples:

d1(u⊗ x) = ux (23.3)d2(x ∧ y) = x⊗ y − y ⊗ x− 1⊗ [x, y] (23.4)

d17→ xy − yx− [x, y] = 0 (23.5)


What is there to check? We should show it’s well-defined on wedge monomials, which isnot hard. We also need to check that dk−1 ◦ dk = 0. When you check this, you get lots oftypes of terms, all of which obviously cancel, either by being sufficiently separated to appeartwice with opposite signs (like in the free resolution of the symmetric polynomial ring), or bysyzygy, or by Jacobi.

We also have to show exactness. This part is for free, by a general principle: if we have acomplex of vector spaces with maps given by matrices (we pick a basis for each term, andthen the matrices depend on the structure coefficients of g with respect to the basis), thenthink about generic t. To with:

General principle: Let F •(t) be a complex of vector spaces, with matrices depending alge-braically on t. Then the H i can only jump for special values. Thus exactness is an opencondition.

Let’s go back and put in the t: we rescale [x, y] to t[x, y]:

dk(x1 ∧ · · · ∧ xk) =∑i

(−1)i−1xi ⊗ (x1 ∧ . . . xi · · · ∧ xk)

− t∑i<j

(−1)j−11⊗ ([xi, xj ] ∧ x1 ∧ . . . xi . . . xj · · · ∧ xk) (23.6)

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This gives the same Lie algebra, just rescaling all the variables: (g, [x, y]t = t[x, y]) ∼= g ift 6= 0. But when t = 0, then the complex is just a complex of polynomial rings, and obviouslyexact.

So it’s exact at t = 0 and hence in an open neighborhood, and we can rescale g to get intoa neighborhood, at least when K is not finite (if it is, replace by its algebraic closure forexample). �

Corollary 23.4.1: Ext•(K,M) is the cohomology of

Mδ1→ HomK(g,M) δ2→ Hom(

∧2(g,M)→ . . . (23.7)

Let g ∈ HomK(∧k−1g,M), then we have

δkg(x1 ∧ · · · ∧ xk) =∑

(−1)i−1xig(x1 ∧ . . . xi · · · ∧ xk)

−∑

(−1)i−j+1g([xi, xj ] ∧ x1 ∧ . . . xi . . . xj · · · ∧ xk) (23.8)

Some examples:

• g = h ∈M , then δ1h(x) = xh.

• g = f : g → M , then δ2f(x ∧ y) = xf(y) − yf(x) − f([x, y]), and the condition to be aone-cocyle is that f be a Lie derivation.


At this point we can see explicitly what a couple Ext groups look like. We have a general machinerydescribing the category of modules over any associative algebra, but in some cases, we can say alot.

24.1 Ext1U(g)(M, N)

We recall that Ext1 classifies extensions. We can calculate is by

0→Mδ1→ HomK(g,M) δ2→ HomK(

∧2g,M)→ . . . (24.1)

Where (δkg)(x1∧· · ·∧xk) =∑

(−1)i−1xi g(x1∧. . . xi · · ·∧xk)−∑i<j(−1)j−i−1g([xi, xj ]∧. . . xi . . . xj · · ·∧

xk).

Let’s calculate Ext1(M,N) = Ext1(K,M∗ ⊗ N) (finite-dimensional, so duals are ok). We recallthat x · φ = xN ◦ φ − φ ◦ xm “= [x, φ]” **describing the action g y Hom(M,N), wherex ∈ g and φ ∈ Hom(M,N) = M∗ ⊗ N**. A 1-cocycle is a map f : g → HomK(M,N) so that0 = δ1f(x∧y) = f([x, y])− (xf(y)−yf(x)). **This mutliplication xf(y) is the action x ·f(y),not the composition x ◦ f(y).**

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So, a 1-cocycle is a map f so that f([x, y]) = “[x, f(y)]− [y, f(x)]”. Let’s say we have an extension0 → N → V → M → 0; this spilts as K-vector spaces: σ : M ↪→ V . How can we make g act onM ⊕N? By

x 7→ñxN f(x)0 xm

ô(24.2)

I.e. f(x) for f ∈ Ext1 are the only possible ways to put something in the upper right corner.

But what if we change the splitting? We had σ : M ↪→ V , and we can change it to σ′(m) =σ(m) + h(m), where h(m) ∈ N . I.e. any other splitting differs by an arbitrary K-linear maph : M → N , and this changes f(x) by xN ◦ h − h ◦ xM = δ1(h). So the 1-cocycles classify thesplitting, and changing the 1-cocycle by a 1-coboundary changes the splitting but not the extension.So Ext1(M,N) classifies extensions up to isomorphism. This completes the proof of the completereducibility.

Of course, the fact that Ext1 classifies extensions is true very generally. We have traced out howthat happens in the case of Lie algebra modules.

24.2 Ext2U(g)(M, N)

A 2-cocycle is g(x ∧ y) ∈M satisfying a condition:

0 = x g(y ∧ z)− y g(x ∧ z) + z g(x ∧ y)− g([x, y] ∧ z) + g([x, z] ∧ y)− g([y, z] ∧ x) (24.3)= x g(y ∧ z)− g([x, y] ∧ z) + cycle permutations (24.4)

A 2-coboundary is g(x ∧ y) such that

g(x ∧ y) = x f(y)− y f(x)− f([x, y]) (24.5)

You can only understand these formulas in hindsight. You can’t look at a formula and say “oh,that’s exactly the formula I need to do this!” Instead, you have some other problem, and theseformulas come up. For example. . .

Consider abelian extensions of Lie algebras 0→ m→ g→ g→ 0 where m is an abelian ideal of g.Well, g y m, but since m is abelian, this action factors through g = g/m. So let’s say we have g

and a g-module m, and we want to understand extensions given this data.

We might have a trivial extension in which σ : g→ g is a g-module splitting, and then g is just thesemidirect product of g with m. But in general we don’t have such a splitting, and so we pick justa K-linear splitting, and let’s measure how far off it is:

As K-spaces, we have g = {σ(x) +m}, and the bracket is

[σ(x) +m,σ(y) + n] = σ([x, y]) + [σ(x), n]− [σ(y),m] + g(x, y) (24.6)

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where g is the error term from being a semidirect extension, and the rest is well-behaved. Theextension is semidirect exactly when there’s a choice of splitting that makes g zero. Definitely g isantisymmetric, since everything else is: g(x, y) = g(x ∧ y).

When will a given antisymmetric error function g(x∧ y) give a valid possible error term? We needto impose a Jacobi identity, so we use the RHS of (24.6) to define the LHS, and test the Jacobiidentity by bracketing with σ(x) + l, and sum over cyclic permutations, hoping to get 0. A bunchof terms die by the usual Jacobi, etc., and the term that survives is exactly the condition that g bea 2-cocycle.

So the 2-cocycles classify extensions of g by m with their splittings, and if we change the splittingby f : g→ m, then g changes by δ2f . So Ext2

U(g)(K,m) classifies abelian extensions 0→ m→ g→g→ 0 up to isomorphism. The element 0 ∈ Ext2 is the semidirect extension g = gnm.

We can use this to understand extensions of semisimple Lie algebras, whence Ext2 vanishes fornon-trivial simple modules.

Theorem 24.1: (Levi)

In characteristic 0, let g be finite dimensional and r = rad(g). Then there exists a semisimplesubalgebra s ⊆ g s.t. g = sn r. s is called the Levi subalgebra.

There is a finer version of this theorem that says that the Levi subalgebras are all conjugate.


Wolog r 6= 0, and we first reduce to the case when r is a minimal non-zero ideal: supposer ) m 6= 0. Well, r/m = rad(g/m), and by induction we get a Levi subalgebra **of g/mwhose lift to g is** s so that s/m ⊆ g/m and s∩r = m and s/m

∼→ g/r **= (g/m)/(r/m)**.Hence m = rad(s) **because its quotient is semisimple and do we know that theintersection of a subalgebra with the radical of an algebra is the radical of thesubalgebra?**, rinse and repeat, and we get s ⊆ s **the Levi subalgebra of s** s.t.s = s ⊕ m **as K-vector spaces** and s ∩ r = 0, hence s

∼→ g/r and we’re done **sincethen s is a Levi subalgbera of g**.

This takes care of the reduction step. So suppose now that r is minimal. Since r is solvable**because it’s a radical**, we have r 6= r′, but by minimality r′ = 0 so r is abelian. Well,Z(g) ⊆ r, so either Z(g) = 0 or Z(g) = r. In the first case, g y r nontrivially, and r isa simple non-trivial g/r module (because the action g y r factors through g/r because r isabelian), and g/r is semisimple **since r is the radical**. By Whitehead’s theorem, incharacteristic 0, Ext2

g/r(K, r) = 0, so 0→ r→ g→ g/r→ 0 is semidirect.

On the other hand, what if Z(g) = r? Then 0 → r → g → g/r → 0 is again an extensionof g/r modules, so it splits **as g/r modules, using Ext1-type arguments, but anyg/Z(g) module is a g-module, but there’s something in this paragraph that I’mmissing**, giving s ∼= g/r an ideal of g, and g is in fact a direct product.

This completes the proof, but we remark that in this second case s ∼= g/r is semisimple, and

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s′ = s = g′, so s is unique. In the other case, s is not unique, but one can say (but we willnot) how flexible it is. �

Recall, we had a correspondence between Lie algebras and Lie groups: we have Lie(−) : f.d. Lie alg/R,C←f.d. Lie groups/R,C, and a map Grp(−) : alg → gp left-adjoint to Lie(−), which gives an equiva-lence of categories onto simply-connected Lie groups. For the proof of this we appealed to Ado’stheorem. But we don’t need to: it follows from Levi’s theorem.

Indeed, the only missing piece is that every g belongs to a G, and by Levi’s theorem g = hn a. Ifa = Lie(A) and h = Lie(H), then since h y a, we have h→ gl(a), and you can check that this liftsto a smooth action H → Aut(A).

Well, if s is semisimple, then its center is 0, and so its adjoint representation is faithful, ands ↪→ gl(s) and by the subgroup theorem, s = Lie(S) for S ⊆ GL(s). And the solvable algebras arebuilt from one-dimensionals, so also have groups.


We finish a few remarks about the representations of Lie algebras, and then move to an intensestudying of the semisimple Lie algebras (and thereby complex and compact Lie groups). That’sthe second half of the course. The end of the first half is Ado’s theorem, which is mystifying butI’ve demystified it, so I’d like to present it. Also, it was in the Lie algebra course I took some timein the last century.

25.1 Ado’s Theorem

We will need a number of facts. One is about how derivations of a Lie algebra act; derivations arelike infinitesimal automorphisms. Anything invariantly attached a Lie algebra should be transferedvia derivations.

The beginning of our story will make sense over any field, but eventually we will move to charac-teristic 0.

Proposition 25.1: Let a be a Lie algebra.

(a) Every derivation of a extends uniquely to a derivation of U(a).

(b) Der a→ DerU(a) is a Lie algebra homomorphism.

(c) If h y a by derivations, then h(U(a)) ⊆ U(a) · h(a) · U(a) the ideal of U(a) generated bythe image of the h action in a.

(d) If N ≤ U(a) is an h-stable two-sided ideal, so is Nn.


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(a) Let d ∈ Der a. For a = Kd ⊕ a; then U(a) ⊆ U(a). The commutative [d,−] preservesU(a) and is the required derivation. Uniqueness is immediate: once you’ve said howsomething acts on the generators, you’ve defined it on the whole algebra.

(b) This is an automatic consequence of the uniqueness: the commutator of two derivationsis a derivation, so if it’s unique, it must be the correct derivation.

(c) LetA1, . . . , Ak ∈ a andH ∈ h. ThenH(A1·Ak) =∑A1 . . . H(Ai) . . . Ak ∈ U(a) h(a)U(a).

(d) Repeat (c), using Ns rather than **missed**

�

Proposition 25.2: Let h y a by derivations, and we form the semidirect product g = hna. Thenthe h-action by derivations and the left (or right) a action on U(a) make a g-action.

In other words, the commutators are what they should be.


We need only check the commutator of h with a. Let u ∈ U(a), H ∈ h, A ∈ a. Then(H ◦ A)u = H(Au) = H(A)u + AH(u) = [H,A]u + AH(u), so [H,A] ∈ g acts as thecommutator of operators H and A on U(a). �

The other thing we need to know is that U(g) is noetherian. Of course, it’s not commutative, butwe mean left-noetherian.

Proposition 25.3: Let U be a filtered algebra (we have subspaces U≤n so that U≤kU≤l ⊆ U≤k+l.If grU is left-noetherian, then so is U . In particular, U(a) is left-noetherian **since grU(a)is a polynomial ring on dim a generators.**


Let I ≤ U be a left ideal. We define I≤n = I ∩ U≤n, and hence I =⋃I≤n. We define

gr I =⊕I≤n/I≤n−1, and this is a left ideal in grU . If I ≤ J , then gr I ≤ gr J , using the fact

that U injects into grU .

So if we have an ascending chain I1 ≤ I2 ≤ . . . , then the grs eventually terminate by assump-tion: gr In = gr In0 for n ≥ n0. But if gr I = gr J , then by induction on n, I≤n = J≤n, and soI = J . Hence the original sequence terminates. �

Our goal is a theorem of Ado:

Goal: Ado’s Theorem: Let K be characteristic-zero. Then every finite-dimensional g has afaithful linear representation g ↪→ gl(V ) **on V f.d.**, and it can be chosen s.t. the largestnilpotent ideal n ≤ g acts nilpotently on V .

Every f.d. Lie algebra can be built by semidirect extensions: if it’s solvable, it’s built from one-dimensional extensions, and the rest is semisimple, and we use Levi’s theorem. So we will similarlybuild the representation step-by-step.

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Proposition 25.4: Zassenhaus’ Extension Lemma

Let g = h n a f.d. and V a f.d. a-module. Suppose that [h, a] ≤ n = largest nilpotency idealof ay V . Then there exists a finite-dimensional g-module W and a surjective a-module mapW � V .

Moreover, this can be done so that the largest nilpotency ideal m ≤ g of W contains n, andif h acts nilpotently on a, then m ⊇ h as well.

Usually this is done by embedding V ↪→W , but that’s an equivalent corollary by considering dualspaces. But the proofs are nicer in this form: the conventional proofs require taking the dual spaceof the universal enveloping algebra.


Consider a Jordan-Holder series 0 = M0 ⊆M1 ⊆ · · · ⊆Mn = V . Then n =⋂

ker(Mi/Mi−1).We define N =

⋂ker(U(a) → End(Mi/Mi−1)). Of course, N ⊇ n ⊇ [h, a]. Then N is an

h-stable two-sided ideal by 25.1 (c), and then so is Nn by 25.1 (d).

Well, Nk is finitely-generated as a U(a)-module (which is Noetherian), hence Nk/Nk+1 isfinitely-generated U(a)/N -module, but U(a)/N is finite-dimensional, so Nk−1/Nk is finitedimensional. This gives us W by induction.

Pick v ∈ V . Well, **for large enough k** Nk kills V , so we get a map a : U(a)/Nn → Vby 1 7→ v, and this is an a-module homomorphism. So pick a basis {vi}di=1; then let

W =d⊕i=1

U(a)/Nn a� V.

U(a) is a g-module, Nn is a-stable (qua left-ideal), and is h-stable; hence is g-stable. SoU(a)/Nn is a g-module.

Well, n acts nilpotently on W , and hence so does the ideal n of g generated by n (MH left thisout of the prepared lecture notes; Exercise). So if h y a nilpotently, then hyW nilpotently,and h + n is an ideal of g. In fact, h + n is an ideal of g. Question from the audience: It’snot generally true that an ideal of h plus an ideal of a is an ideal of h n a. But it is by theassumption that [h, a] ⊆ n. Answer: Aha. That’s in fact why n is an ideal of g, too. And itdoes say so in my notes.

So, refresh, n an ideal of g acts nilpotently on W , and h + n is an ideal of g, and we’d like toshow that it acts nilpotently on g. Well, n ⊆ h + n. So we finish with a

Lemma 25.5: If g = h + n where h is a subalgebra and n is an ideal, and if h and n both actnilpotently on V , then g does.


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Wolog V 6= 0. By Engel’s theorem, we can take v ∈ V n 6= 0. If H ∈ h and X ∈ n, then

XHv = [X,H]︸︷︷︸∈n

v +H Xv︸︷︷︸=0

= 0,

so hv ∈ V n. So we’ve found an invariant vector for g y V , and iterating gives thatg y V nilpotently. �

�

We’re almost ready for Ado’s theorem. We’ll need one more lemma, and will reserve the proof ofAdo for next time.

Lemma 25.6: If r is solvable and d y r is a derivation, then d(r) ⊆ the largest nilpotent idealn(r). In particular, if g = hn r, then [h, r] ⊆ [g, r] ⊆ n(r).


Let t = hdn r. Then t′ is nilpotent, by Lie’s theorem — we are using characteristic-zero here— and d is ad d in t′: d(r) = [d, r] ⊆ t′. But r ⊇ t′. Of course, t′ is a nilpotent ideal. �


We want to start off by finishing off the proof of Ado’s theorem. Then we move to the second halfof the course: the structure and classification of semisimple Lie algberas.

We had a number of results last time:

Lemma 26.1: (Zassenhaus)

Let g = h n a finite-dimensional, and V a finite-dimensional a-module. Let n ≤ a be thelargest nilpotency ideal of a y V . Suppose [h, a] ⊆ n. Then there exists a finite-dimensionalg-module W with an a-module surjection W � V such that the largest nilpotency idealm ≤ g yW contains n, and also if h y a nilpotently, then h ≤ m too.

The proof was by induction, using free powers of nilpotentcy ideals.

Lemma 26.2: (In characteristic zero,) let r be a solvable Lie algebra and n(r) its largest nilpotentideal. Then every derivation of r has image in n(r). In particular, if g = sn r — indeed, if r

is any ideal in some larger g **why?** — then [g, r] ≤ n(r).

Theorem 26.3: (Ado)

Let K be characteristic zero, and g a finite-dimensional Lie algebra. Then g has a faithfulrepresentation g ↪→ gl(V ) and it can be chosen so that the largest nilpotent ideal n ≤ g actsnilpotently.

The meat of the proof is in the first lemma:

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We induct on dim g. The g = 0 case is trivial.

Case I: g = n is nilpotent. Then g 6= g′, and so we choose a subspace a ⊇ g′ of codimen-sional 1 in g. This is automatically an ideal, and we pick x 6∈ a, and h = 〈x〉. Anyone-dimensional subspace is a subalgebra, and g = hn a. By induction, a has a faithfulmodule V ′ and acts nilpotently.

The hypotheses of 26.1 are satisfied, and we get an a-module homomorphism W � V ′

with g y W nilpotently. As yet, this might not be a faithful representation of g, butcertainly a acts faithfully on W because it does on V ′. But x might kill W . So let’s pick

a nilpotent g/a = K-module W1, e.g. t 7→ñ

0 t0 0

ô. Then V = W ⊕W1 has a faithful

nilpotent g representation. **Better notation is to use x rather than t above?**

Case II: g is solvable but not nilpotent. Then g 6= n ⊆ g′. We pick a codimension 1such that n ⊆ a, and x and h = Kx as before. Then g = h n a. Then n(a) ⊇ n

**why?** and we have a y V ′ faithfully by induction, with n(a) y V ′ nilpotently.Then [h, a] y nilpotently, so we use 26.1 and get g yW �

aV ′ and n yW nilpotently.

We form V = W ⊕W1 as before. n being contained in a acts as 0 on W1 and so stillnilpotently on V , and g y V is faithful.

Case III: general. By Levi’s 24.1, there is a splitting g = s n r with s semisimple and r

solvable. By Case II, we have a faithful r-module representation V ′ with n(r) y V ′

nilpotently. By 26.2 the conditions of 26.1 apply, so we have g y W �rV ′, and since

n ≤ r we have n ≤ n(r) so n y nilpotently. We want to get a faithful representation,and we need to make sure it is on s. But s = g/r is semisimple, so has no center, soad : s y s is faithful. So we let W1 = s = g/r as g-modules, and g y V = W ⊕W1 isfaithful with n acting as 0 on W1 and nilpotently on W .

�

So we see that Ado’s theorem is really a corollary of Levi’s theorem. We make one remark: theproof technique is that we can assemble g out of smaller pieces, and we discuss one corollary of theproof.

Let g be a Lie algbera over R or C, and g = h n a, which corresponds to G = H n A. Indeed,whenever g→ gl(V ) = LieGL(V ), we get a map G→ GL(V ). So all representations of Lie algebrasare actually representations of Lie groups. So we can follow the above construction, and prove thatwe get a representation of any simply-connected Lie group. Indeed, we can arrange for it to befaithful: the representations on W are faithful by induction, and W1 takes some care, but each timewe’re either extending by a one-dimensional of a semisimple. Certainly R and C have faithful linearrepresentations. We will completely classify the structure of semisimple Lie groups, and show thatsemisimple Lie groups have faithful linear representations.

Hence, we will conclude that every simply-connected Lie group has a faithful finite-dimensional

77

representation. Question from the audience: What about SO(2,R) **or SL? not sure**?Answer: Maybe we want to add the word “complex”.

Maybe we will discuss this in the next set of exercises. By the way, there is another set of exer-cises on the website, in case you haven’t noticed. If you noticed too soon, notice again: they’vechanged.

This concludes our general discussion of Lie groups, We will move to the semisimple group, andit will turn out that the representations of complex, compact, etc. groups are all classified by thesame combinatorial data.

So that’s the plan, but much work needs to be done: inside each semisimple Lie algebra we needto find a fair amount of structure. But we should be motived by some examples.

26.1 Semisimple Lie algebras

From now on, it will be a blanket assumption that all Lie algebras will be finite-dimensional over afield of characteristic 0. Eventually we will restrict to R or C, but actually everything we do overC will hold over any algebraically closed field.

A Lie algebra g is reductive if (g, ad) is completely reducible. Then g =⊕

ai, and these are idealsand hence [ai, aj ] ⊆ ai∩aj = 0, so g =

∏ai as Lie algebras. And each ai is either simple non-abelian

or one-dimensional. So g is reductive iff it is a semisimple × an abelian. Then the abelian factoris the center, and the semisimple bit is the derived subalgebra.

So if we can produce a reductive Lie algebra with no center, then it is obviously semisimple. Thiswill be useful, since up to now the only semisimple Lie algebra we have is sl(2), which is semisimple(and hence simple, since we know all the one- and two-dimensional Lie algebras) by computing theKilling form.

But we also have the classical Lie algebras. We can be a little loose: g/R is semisimple iff C ⊗ g

is. The complex semisimple algebras are on the following list (plus finitely many exceptionalones):

• sl(n,C) = {X ∈ gl(n) s.t. trX = 0}

• so(n,C) = {X ∈ gl(n) s.t. X +XT = 0}

• sp(2n,C) = {X ∈ gl(2n) s.t. JX + XTJ = 0} where J =ñ

0 In−In 0

ô; we remark that

J2 = −I2n.

We define **the Hermitian conjugate** X∗ = XT .

Claim: Ant ∗-closed subalgbera of g = gl(n,C) is reductive.

Proof of Claim:

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Indeed, we define a pairing (, ) on gl(n,C) by (X,Y ) = < tr(XY ∗), which is a positive-definiteR-valued symmetric bilinear form. Indeed, (X,X) =

∑ |Xij |2. It is invariant, in the sensethat ([Z,X], Y ) = −(X, [Z∗, Y ]). **So [Z∗,−] is adjoint to [−, Z].**

So if we have an ideal a ≤ g ⊆ gl(n,C), the invariance says that a⊥ ⊆ g∗ **the Lie algebraof Hermitian conjugates** is an ideal. Of course, if g = g∗, then the perp of any ideal isanother ideal. And by positive-definiteness, g = a ⊕ a⊥. So keep breaking down the idealseventually into irreducibles. �

And all the classical Lie algebras are more-or-less obviously closed under Hermitian conjugation.The only one that takes checking is the last: JX+XTJ = 0 implies X∗J+JXT∗ = 0, so conjugateby J and get that JX∗ +X∗TJ = 0.

So the classical algebras are all reductive, and almost all are semisimple. A few exceptions: so(2,C)is abelian. so(4,C) ∼= so(2) × so(2). But for n > 4 we have so(n,C) has zero center. So too dosp(2n,C) and sl(n,C) for n > 1. So they’re all semisimple, and in fact except for the small n theclassical groups are all simple.

We will call sl(n) “An−1”. The sos behave rather differently for n even or odd: the odd ones behavemore like the simplectic algberas. We will call Bn = so(2n+1) and Dn = so(2n). And sp(2n) = Cn.We will have five exceptional simples: E6, E7, E8, F4, and G2. These are all the simples over C,and next time we will describe a few of these in detail.


We begin our study of semisimple Lie algebras. From here, we will move to compact Lie groups,complex Lie groups, algebraic groups, etc.

First, we take a close look at two of the classical groups, developing combinatorial data. Then wewill build a procedure from first principles to extract this combinatorial data.

27.1 sln and sp2n over C

For each of the algebras, we would like to extract an abelian subalgebra: For sln, we use

hdef=

z1

. . .zn

s.t.∑

zi = 0

(27.1)

The dimension of sln is n2 − 1 = 2(n

2

)+ n− 1

For sp2ndef= {x ∈ gl2n s.t. Jx+ xTJ = 0}, it will be helpful to redefine J . We can use any J which

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defined a non-degenerate antisymmetric bilinear form, and we take:

J =

1

. ..

1−1

. ..

−1

(27.2)

then sp2n =®ñ

A B = BR

C = CR −ARô´

(27.3)

where AR is A reflected across the antidiagonal. The dimension is clearly 2n2 + n.

Then we can take as our abelian subalgebra

h =

z1

. . .zn−zn

. . .−z1

(27.4)

In both cases, ad h is diagonalizable. In sln, we have [H,Eij ] = (zi − zj)Eij when i 6= j, and so theroots are {0} ∪ {zi − zj : i 6= j}.

**I missed the board’s computation of the action ad : h y sp2n. But the point is that thenatural basis in 27.3 diagonalizes the action.** The roots of sp2n are {0}∪{±2zi}∪{±zi±zj :i 6= j}.

The roots break g = sp or sl into eigenspaces:

g = h⊕⊕

α a root

gα (27.5)

where h = g0 and in fact the root spaces gα are one-dimensional for α 6= 0. This decomposition isreally good: [gα, gβ] ⊆ gα+β.

Moreover, since each is one dimensional, by taking the essentially unique guys in gα and g−α andconjugating them we get in h; let’s call the bracket [gα, g−α] = hα. Then gα ⊕ g−α ⊕ 〈hα〉 is asubalgebra, and in fact is isomorphic to sl2, since α(hα) 6= 0, **and the bracket is [g±α, hα] =±αgα and [gα, g−α] = αhα, maybe?**. For example **missed, finding the sl2s in sln**.

We remark that throughout this discussion, the zis are linear functionals on h, so the αs are vectorsin the dual space to h.

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In sp, the sl2s are a little more complicated. **This one I didn’t miss, but keeping up withthe board with matrices and anti-diagonals is sufficiently hard, and I haven’t had anycoffee today.**

Let’s draw the rank-2 picture. The rank is the dimension of h. So this case is sl3: there’s α1 = z1−z2,α2 = z2 − z3, and α1 + α2 = z1 − z3. Then we get a picture in the dual space to sl2, and if wechoose an inner product in which the zs are orthonormal, then these three vectors generate a perfecthexagon:

• 0

•α3 = α1 + α2

•α1

•α2

•−α3

•−α1

•−α2

The picture for sp4 we have α2 + 2α1 = 2z1, α2 + α1 = z1 + z2, α1 = z1 − z2, and α2 = 2z2. Wehave a diamond:

• 0

•α2 + 2α1

•α1

•α2 + α1

•α2

•

•

•

•

There’s a lot of symmetries. In sl2, the group is S3, and in sp4, the group is the signed permutationgroup.

Let’s generalize this. We pick an element of h on which none of the αs are zero. This divides the rootsinto some positive roots. For example, we take as our choice anything with z1 > z2 > · · · > zn > 0.Then in sln, the positive roots are zi − zj for i < j, but the minimal ones are α1 = z1 − z2,α2 = z2−z3, . . . , αn−1 = zn−1−zn. The minimal roots are a basis of h∗ the dual space to h. Theseminimal roots we call the simple roots, and then any positive root zi− zj = αi +αi+1 + · · ·+αj−1.

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We can make a partial order on the positive roots by looking at these sums:

•α1

•α2

•α3

. . . •αn−1

•α1 + α2 • • . . . •αn−2 + αn−1

• • • •• • •• ••θ = z1 − zn = α1 + · · ·+ αn−1

��??? ��

??? ��??? ��

???��

??? ��??? ��

?? ��???

��??? ��

??? ��???

��??? ��

???��

???

Question from the audience: What does “simple” mean? Answer: The minimal ones. Ques-tion from the audience: That depends on a choice? Answer: Yes, but there’s a symmetry.They Weyl group will take any system of simple roots to another system. We pick an element of h

to separate the roots into positive and negative ones, and then the simple roots are positive rootsthat are not the sum of two positive roots.

It will turn out by the general theory that we can always do this, and the simple roots will alwaysbe a basis of h∗ that Z-generate all the roots and N-generate the positive ones.

In sp2n, we can take the same decreasing h ∈ h, and then the positive roots are zi − zj , zi + zj ,and 2zi. The minimal ones are are αi = zi − zi+1 for 1 ≤ i ≤ n − 1 and αn = 2zn. Then2zi = 2(αi + · · ·+ αn−1) + αn, and zi + zj = αi + · · ·+ αj−1 + 2(αj + · · ·+ αn−1) + αn.

The picture **reconstructed from memory and minimal notes; I think I might have itbackwards**:

•α1

•α2

. . . •αn−1

•αn

• • . . . • •• • • • •• • • •• • • •• • •• • •• •• •••θ = 2(α1 + · · ·+ αn−1) + αn

��??? ��

??? ��??? ��

???��

??? ��?? ��

??? ��??? ��

��??? ��

??? ��??? ��

???��

??? ��??? ��

??? ��

��??? ��

??? ��???

�� ???�� ???

��

??? ��???

�� ???��

��???

��


**I was ten minutes late, on account of my computer not being charged.**

**Copied from the board, presumably in the case when g = sl or sp, given that it’sbased on the root pictures we worked out last time:** Every x ∈ g has xθ in the ideal thatit generates. But xθ generates g.

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How much data do you need to do this style of combinatorial argument? It depends only on theCartan matrix A, given by Aij = αi(hαj ). For sln, we have αi = zi− zi+1 and hαj = εj − εj+1, andziεj = δij , so the Cartan matrix for sln is

2 −1 0 . . . 0

−1 2 −1...

0 −1. . . . . . 0

.... . . 2 −1

0 . . . 0 −1 2

(28.1)

For spn, we have αn−1 = zn−1 − zn and αn = 2zn, and hαn−1 = εn−1 − εn and hαn = εn. So thematrix for sp2n is almost the same, except has a −2 in the lower corner where there should be a−1:

2 −1 0 . . . 0 0

−1 2 −1...

...

0 −1. . . . . . 0

......

. . . 2 −1 00 . . . 0 −1 2 −10 . . . . . . 0 −2 2

(28.2)

We draw Dynkin diagrams: a node for each vertex; i and j are not connected if Aij = 0; singly

connected if Aij is a blockñ

2 −1−1 2

ô, and a double arrow towards the shorter root when the

block isñ

2 −1−2 2

ô. It turns out that our diagrams will be

• An = sln+1: • • . . . •

• Bn = so2n+1, for n ≥ 3: • • . . . • •+3

• Cn = sp2n for n ≥ 2: • • . . . • •ks

• Dn = so2n for n ≥ 4: • • . . . ••

•

��???

• G2: • •_*4 . The triple arrow means the block in the Cartan matrix isñ

2 −1−3 2

ô.

• E6,7,8: • • • . . . ••

Seeing these is a good exercise.

We also introduce the Weyl group, which is the group of symmetries of the root system. For sln,R r {0} = {zi − zj : i 6= j}, and Sn y R r {0} by permuting these roots **perhaps we mean

83

sln+1**. This is generated by the reflections (i, i+ 1), which act by αi 7→ −αi. We define the dualvectors α∨i by 〈α∨i , α〉 = α(hαi); then the reflection (i, i+ 1) sends α 7→ α− 〈α∨i , α〉αi.

For sp2n, the Weyl group is generated by the symmetric group (switching positions as with sln)and the sign change αn 7→ −αn, hence we get all sign changes. This is the “hyperoctahedral group”Bn = Sn n (Z/2)n. The Weyl group in sl acts transitively on the roots; in sp there are two classesof roots, long and short, and it is transitive on each class. It’s also transitive on the choices ofpositivity.

Where does the Weyl group live? There are Lie groups SL(n,C) and Sp(2n,C) associated tothese algebras. You might think that the Weyl group lives in these, but it doesn’t quite, e.g.some permutations have determinant −1. What will actually happen is that associated to h wewill have a group T = (C∗)n−1, “T” for “torus”. And we look at the normalizer N(T ) in SL(n),and the normalizer N(T )/T can’t see the signs of the permutations, and in particular is the Weylgroup.

We will soon return to general theory.

We won’t start with the Cartan matrix — it’s rather abstract — but with the observation that inthe examples, we saw that gα⊕g−α⊕〈hα〉 ∼= sl2. The fact that there are so many sl2s makes g intoan sl2 module in lots of ways, and the representation theory of sl2 is particularly easy. And thisrepresentation theory entirely controls the underlying combinatorics. We will eventually classifyall the representations of all the classical algebras, and this also is based on the sl2 structure. Thephilosophy is that sl2 governs everything.

28.1 sl2(C) modules

We let C be any algebraically closed field of characteristic 0.

We’ve seen alread that sl2(C) is (semi)simple, hence finite-dimensional modules are direct sums ofirreducibles. So we would like to find the irreducibles, and be able to look at any other module andtell its decomposition.

Of course, sl2 has a tautological representation on C2, where E 7→ñ

0 10 0

ô, F 7→

ñ0 01 0

ô, and

H 7→ñ

1 00 −1

ô. A better picture:

•v0

•v1

Hyy

Hyy

F

E

II

The advantage of these pictures is that as the representations get larger, they’re easier to draw**except if you don’t know XY well**.

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Let’s think about how SL(2) y C2, and understand the sl2 action as the infinitesimal version ofthis action. Then the two-dimensional representation will be the rep on the linear functions on C2,but we can also act on polynomials of any degree.

(exp(−tE))ñxy

ô=ñ

1 −t0 1

ô ñxy

ô=ñx− tyy

ôd

dt

∣∣∣∣t=0

exp(−tE) =®ñ

xy

ô7→ñ−y0

ô´d

dt

∣∣∣∣t=0

exp(−tF ) =®ñ

xy

ô7→ñ

0−x

ô´Similarly for H. Overall we see that

E = −y∂x, F = −x∂y, H = −x∂x + y∂y (28.3)

Now, we can notice that for each of these operators, the total degree in x and y is zero (since∂x has degree −1 in x): they’re all homogeneous. I.e. they preserve the degree of a polyno-mial, i.e. Sn(C2) = {homogeneous polynomials of degree n in x and y} is a submodule of SL(2) y{functions}.

Let’s understand these submodules. We have videf=(ni

)xiyn−i for i ∈ {0, . . . , n}. The reason for

the binomial coefficients is ultimately because of the quantum groups, so take Kolya’s class nextsemester, but we’ll leave it at the fact that d

dxxn

n! = xn−1

(n−1)! . Then E : vi+1 7→ (n− i)vi and kills v0;F : vi−1 7→ ivi and kills vn, and H : vi 7→ (n− 2i)vi:

•v0=yn

•v1=nxyn−1

•v2=(n

2

)x2yn−2

...

•vn−1=( nn−1

)xn−1y

•vn=xn

H=nyy

H=n−2zz

H=n−4zz

{{

H=2−nzz

H=−nzz

F=1��

n=E

HH

F=2��

n−1=E

HH

HH

��

JJ

F=n��

1=E

HH

If we didn’t have the binomial coefficients, then the coefficients on E and F would be backwards.We point out that Sn(C2) is thus an irreducible representation of SL(2). Why? Because if you

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start at any linear combination of dots, then applying E enough times leaves you just at the topdot v0, and v0 generates everything.

Lecture 29 November 3, 2008

We recall from last time our picture of Vn, a (n + 1)-dimensional irreducible representation ofsl(2,C).

•v0

•v1

•v2

...

•vn−1

•vn

H=nyy

H=n−2zz

H=n−4zz

{{

H=2−nzz

H=−nzz

F=1��

n=E

HH

F=2��

n−1=E

HH

HH

��

JJ

F=n��

1=E

HH

(29.1)

Let V be any finite-dimensional irreducible sl(2,C) module. Suppose v ∈ V is an H eigenvectorwith Hv = λv. Then HEv = [H,E]v+EHv = (λ+2)Ev, and similarly Fv is an H-eigenvector. Sothe space spanned by H-eigenvectors is a submodule, and since V is irreducible, H acts diagonally.By finite-dimensionality, there’s a vector v0 with Hv0 = λ0v0 and Ev0 = 0, and by PBW 12.1,{F kElHm} spans U(sl2), so vi = F iv0/i! are a basis of V . There is a last non-zero vn = Fnv0/n!with Fvn = 0.

Checking commutators, we have Ev1 = EFv0 − ([E,F ] + FE)v0 = λ0v0, and by induction Evk =(λ− k + 1)vk−1; thus we must have λ0 = n, and V is our representation Vn above (29.1).

29.1 Generalizations

One of the things we wanted to do was to act diagonally. In the general case this is too much to ask,but we can get generalized eigenspaces. In fact, this is remarkable: if a bunch of endomorphismscommute, then they have generalized eigenspaces, but in fact it’s enough for the algebra to benilpotent. This is the missing theorem between Lie’s theorem and Engle’s theorem.

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Goal: If h is a nilpotent Lie algebra acting on finite-dimensional V , then we get joint generalizedeigenspaces V =

⊕Vλ where λ : h → K is a linear functions and Vλ = {v ∈ V s.t. ∀H ∈

h, ∃n s.t. (H − λ(H))nv = 0}.

For an arbitrary Lie algebra, it’s not clear that you’d have any such vectors at all.

Lemma 29.1: Let Vλ,H = {v s.t. ∃n s.t. (H − λ)nv = 0}, where H ∈ h nilpotent, λ ∈ K, and V afinite-dimensional h-submodule. Then Vλ,H is an h-submodule.

Question from the audience: Any field? Answer: We won’t need any assumptions on K forthis lemma, but Vλ,H might be 0 if K is not algebraically closed.


adH is nilpotent, so let h(m)def= ker(adH)m, which is h when m ≫ 0. We will show

that h(m)Vλ,H ⊆ Vλ,H by induction, the case h(0) being trivial. Well, let Y ∈ h(m). Then[H − λ, Y ] = [H,Y ] ∈ h(m−1), and so in their action on V , [(H − λ)n, Y ] =

∑k+l=n−1(H −

λ)k[H,Y ](H − λ)l, and for v ∈ Vλ−H , we have that (H − λ)nv = 0 for n≫ 0. We want tostudy (H − λ)nY v = Y (H − λ)nv + [(H − λ)n, Y ] v, which is just (H − λ)k[H,Y ](H − λ)lvfor n large, but the only surviving terms have either l or k large. If l is large, then this killsv; by induction, [H,Y = (H − λ)lv ∈ Vλ,H , and so when k is large, this term also dies. �

Corollary 29.1.1: Vλ =⋂H∈h Vλ(H),H is also an h-submodule.

Now assume that K is algebraically closed.

Proposition 29.2: V =⊕Vλ.


Let H1, . . . ,Hk ∈ h, and Wdef=⋂i Vλ(Hi),Hi . Then we claim that W =

⋂Vλ(H),H where H

ranges over the span 〈H1, . . . ,Hk〉 of the His. This is by Lie’s theorem, by writing everythingin a basis in which it’s upper-triangular.

Well, W is a submodule, so pick Hk+1 6∈ 〈H1, . . . ,Hk〉; we can decompose W into generalizedeigenspaces of Hk+1, rinse and repeat. �

The generalized eigenspace Vλ is called a weight space.

Note: because H(v ⊗ w) = Hv ⊗ w + v ⊗Hw, we see that if v and w are generalized eigenvectorswith eigenvalues λ and µ, then v ⊗ w has generalized eigenvalue λ + µ. I.e. The weight spaces ofV ⊗W are (V ⊗W )λ =

⊕α+β=λ Vα ⊗Wβ.

Corollary 29.2.1: If h ≤ g and h is nilpotent, then the weight spaces of (g, ad) satisfy [gα, gβ] ≤gα+β.

Proposition 29.3: Let h ≤ g be a nilpotent subalgebra. Then the following are equivalent:

(a) h = N(h) def= {X ∈ g s.t. [X, h] ⊆ h}

(b) h = g0

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Such an h is a Cartan subalgebra of g.


Let N (i) def= {X ∈ g s.t. (ad h)iX ⊆ h}. Then N (0) = h ⊆ N (1) = N(h) ⊆ N (2) ⊆ . . . .By finite-dimensionality they eventually stabilize, but in fact they eventually stabilize ong0. So (b) implies (a), but each term is just determined by the previous, so if (a), thenN (i) = N (i+1) = g0 and (a) implies (b). �

Proposition 29.4: Every finite-dimensional g over infinite K has a Cartan subalgebra.


We want to consider those elements that are as diagonal as possible, and thus the weightspace decomposition is as fine as possible, and the 0-weight space is as small as possible. Wesay that X ∈ g is regular is g0,X has minimal dimension. We claim that if X is regular, thenh = g0,X is a Cartan subalgebra.

Why is it a subalgebra (not in MH’s notes)? “It’s easy to check,” because of the weight-spaceadding. We want to show first that it’s nilpotent. Suppose not — this is subtle, because there’ssome algebraic geometry going on here. Let U def= {H ∈ h s.t. adH|h is not nilpotent}. Wecannot have U = 0 if h is not nilpotent. But U = {H s.t. (adH|h)d 6= 0}. This is Zariski-open: it’s the complement of the set of matrices for which the polynomial (in the coefficients)(adH|h)d vanishes. We define V def= {H ∈ h s.t. H acts invertibly on g/h}. This is also non-zero, since we’ve modded out by eigenvalues being 0. This is also Zariski-open, again bytaking bases.

Question from the audience: Can you repeat what is the Zariski topology? Answer:Zariski-closed sets are the zero-sets of (systems of) polynomials.

The point is that any two non-empty Zariski-open sets of a vector space **over infiniteK** have non-empty intersection: U ∩ V 6= 0.

Next time we will explain this and finish the proof.

�


Today we will pick up in the middle of the proof we left last time. First, a little algebraic geometryabout Zariski closed sets. We say that X ⊂ Kn is Zariski closed if X = {~x s.t. pi(~x) = 0∀u} forsome polynomials p1, . . . . (Possibly infinitely many.) The complement of a Zariski closed set isZariski open, and these are the open sets of a topology.

A few remarks. If K is infinite, and U and V are Zariski open and non-empty, then U ∩ V 6= ∅.Because let u ∈ U and v ∈ V , and consider the line L passing through u and v. Then L ∩ U and

88

L ∩ V are each finite sets, as they are the loci of polynomials, hence L contains infinitely manypoints in U ∩ V .

Moreover, let K = C. Then if U is Zariski open, then it is path-connected. Indeed, take two pointspoints and their connecting (complex) line L. Then L ∩ U ∼= Cr {finite} is path-connected.

30.1 Continuations of last time

We define a Cartan subalgebra h ≤ g to be a nilpotent subalgebra such that h = Ng(H). Equiva-lently, h = g0,h.

Proposition 30.1: Let K be infinite. Then a Cartan subalgebra exists.


Say X ∈ g is regular if dim g0,X is minimal. Question from the audience: What is g0,X?Answer: If g y V , we define Vλ,X = {v s.t. (X − λ)nv = 0 for some n}. This is that forad : g y g In any case, let h = g0,X , which is a subalgebra by weight additivity. We claim

that h is nilpotent. Suppose it is not; then ∅ 6= Udef= {H ∈ h s.t. adhH is not nilpotent} =

{H s.t. (adH)dim h 6= 0} is Zariski open. Moreover, h qua subalgebra is a submodule ofad : h y g. We ket V def= {H ∈ h s.t. H acts invertibly on g/h}. This is also nonempty,because X ∈ V , and also Zariski-open (a matrix is invertible iff its determinant is non-zero,and the determinant is a polynomial in the matrix entries). So, there exists Y ∈ U ∩V . ThenadY preserves all gα,X since Y ∈ g0,X , and invertible on all with α 6= 0. Then g0,Y ⊆ g0,X = h,but since Y ∈ U , g0,Y 6= h. This contradicts minimality; hence h is nilpotent.

Since h is nilpotent, it has a weight-space decomposition. We consider the generalizedeigenspace g0,h of its action on g. Then h ⊆ g0,h ⊆ g0,X = h. **MH said why eachof these, but I was a sentence behind. First is since h is nilpotent, so acts onitself with generalized eigenvalue 0, and the second is since X is in h, and the lastis by definition.** �

Question from the audience: Is the proposition actually false for finite fields? Answer: Prob-ably. But the goal of all this is to classify semisimple Lie algebras over C.

So, we take a semi-simple Lie algebra, find a Cartan subalgebra; we’ll see that in fact the Cartanacts diagonalizably. We will get some combinatorial data, but we’d like to know that this datadoesn’t depend on our choice of Cartan.

Proposition 30.2: When K = C, all Cartan subalgebras are conjugate by automorphism of g.

Question from the audience: When you say C, are you going to use analysis? Answer: Yes, Ireally mean C. We need the analysis to get the automorphisms; we need to use the theory of Liegroups. We will also need the path-connectedness of Zariski opens, so this doesn’t even work overR.

89


We consider ad : g → gl(g). In particular, ad g ⊆ gl(g) is a Lie subalgebra, and there is acorresponding connected Lie subgroup Int g of GL(g) generated by exp(ad g). But since g

acts by derivations, exp(ad g) acts by automorphisms, hecne Int g ⊆ Aut(g).

Let h be a Cartan, and g =⊕

gα where gα = gα,h, and in particular h = g0. Since g is

finite-dimensional, we have a non-empty set Rhdef= {H ∈ h s.t. α(H) 6= 0∀α 6= 0} = {H ∈

h s.t. g0,H = h}.

We want to consider the orbits of the action σ : Int g × Rh → g. Let’s pick a Y ∈ Rh andlook at the tangent space at (e, Y ). What is dσ(T(e,Y ) Int g × Rh)? Since σ(e, Y ) = Y , weidentify TY g = g. Well, if we vary the first component, by construction we have an actionby conjugation: X 7→ [X,Y ], and so the image is adY (g). But Y acts invertibly, henceadY (g) ⊇ ⊕

α 6=0 gα, and varying the second coordinate (in all h directions), we see thatdσ(T(e,Y ) Int g × Rh) also contains g0 = h. Hence dσ(T(e,Y ) Int g × Rh) = g = TY g. Thus,(Int g)(Rh) contains a neighborhood of Y , and hence is open.

Now, let’s understand how big is the generalized nullspace. We let Y ∈ g range, and considerg0,Y . The size of the generalized nullspace depends on the characteristic polynomial, andthe coefficients of the characteristic polynomial depend polynomially on the matrix entriesof adY . The condition that dim g0,Y ≥ r is that the last r coefficients of the characteristicpolynomial of adY are 0. So {Y s.t. dim g0,Y ≥ r} is Zariski closed, and dim g0,Y is upper-semicontinuous. Or possibly lower-semicontinuous (in the Zariski topology, but then also inthe normal topology, since Zariski closed sets are closed); this is in the category of things likeleft versus right cosets that I can never remember.

In any case, dim g0,Y is constant on (Int g)Rh, because it equals dim h. Thus this is theminimal value on the closure of (Int g)Rh. If we knew that (Int g)Rh was Zariski open, we’dbe happy: then dim h is the minimal value and (Int g)Rh are all regular sets. In fact, we arehappy: since dim g0,Y is upper-semicontinuous in Zariski, its minimal locus Reg is Zariski-open, hence dense, so it meets (Int g)Rh and so dim h is the minimal dimension. Rh ⊆ Reg.Conversely, the union over all Cartans is the regular elements:

⋃h CartanRh = Reg. And the

unions of the “orbits” (Int g)Rh is all the regular elements.

But (Int g)Rh, as it is a connected group acting on a connected set (Cn minus some hyper-planes), and Reg is Zariski-open and hence connected. But the different images of (Int g)Rh

are disjoint, and their union is Reg, so one of them is all of them.

Reviewing, h is Cartan so contains regular elements, and any other regular is the image underInt g of a regular in h. This implies that every Cartan is in (Int g)h. �

In the last few minutes, we begin to consider the Jordan decomposition of elements of a semisimplealgebra.

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30.2 Jordan decomposition

Let X ∈ End(V ). Then X = Xs + Xn has a uniquely decomposition with Xn nilpotent and Xs

diagonal. This is over algebraically closed.

A semisimple Lie algebra has the property that its adjoint representation is faithful. We want tosay something stronger:

Lemma 30.3: Let K be algebraically closed. Then Der g contains Xs and Xn forall X ∈ Der g.


If X ∈ Der g, it induces a weight-space decomposition⊕

λ gλ of generalized eigenspaces ofX, and the fact that it’s a derivation implies that the weights add: [gµ, gν ] ⊆ gµ+ν . But Ywhich acts as λ on gλ commutes with X, is diagonal, and X − Y is nilpotent, because all itseigenvalues are 0. So Y = Xs, but this is a derivation, as is the difference Xn = X −Xs. �

Corollary 30.3.1: If g is semisimple, then every X ∈ g has a unique Jordan decomposition X =Xs +Xn such that [Xs, Xn] = 0, adXs is semisimple, and adXn is nilpotent.


ad : g → Der g is injective because Z(g) = 0 by semisimplicity, and is surjective, because allderivations are inner, because Der / ad g = Ext1(g,K) for any g (HW), which is 0 when g issemisimple. �

**We are now being kicked out of the room.**


**I missed the first 10 minutes.**

Theorem 31.1: Schur’s Lemma (stronger version for algebraically closed field)

Let K be algebraically closed, and V an irreducible finite-dimensional module. Then EndV =K.


Let φ ∈ EndV and λ an eigenvalue. Then φ− λ is singular, hence 0. �

Proposition 31.2: Let g be semisimple, K algebraically closed. In every finite-dimensional g

module σ : g→ gl(V ), we have σ(X)s = σ(Xs) and σ(X)n = σ(Xn).

Here we define σ(X)s,n as the semisimple and nilpotent parts of the matrix σ(X) ∈ gl(V ). xs,n wedefined last time in terms of the isomorphism ad : g

∼→ Der g (every derivation is inner, becauseExt1(g,K) = 0 from the HW).

The proposition will imply that any Cartan subalgebra acts diagonalizably on any irreducible.

91

**That was copied from the board. The proof of the proposition is from Alex Fink. Ipick up live at the end of the proof of the proposition.**


We can reduce to the case when V is irreducible. SInce g is semisimple, g =∏

gi with eachgi simple, and if g y V irreducibly, then gi y V as 0 for all but one i; the last one is faithful.So replace g by that gi, and we have σ : g ↪→ gl(V ) with g simple.

It’s enough to show that σ(X)s is in σ(g): then σ(X)s = σ(s), σ(X)n = σ(X)−σ(X)s = σ(n),and s, n commute and sum to X; then they act semisimply and nilpotently respectively inthe adjoint action, ad : g y g is a submodule of g y gl(V ).

Since g = g′ by semisimplicity, everything in g is a commutator, so g is a subalgebra of sl(V ).By Schur’s lemma 31.1, the centralizer of g in gl(V ) consists of scalars. So the centralizer insl(V ) is 0, because we’re in characteristic 0.

We consider the normalizer n = {X ∈ sl(V ) s.t. [X, g] ⊆ g}. **MH uses “N”, but it’s aLie algbera. But, of course, “n” looks like a nilpotent algebra.** So n acts faithfullyon g (because the centralizer of g in sl is 0) by derivations (since its action is by a bracket).But all derivations are inner, so n y g factors through g y g, but it’s faithful, so n = g.

So it suffices to show that σ(X)s ∈ n. Well, σ(X)n ∈ sl(V ) because it’s nilpotent, soσ(X)s ∈ sl(V ) too. We find a generalized eigenspace decomposition of V with respect toσ(X): V =

⊕Vλ. Then σ(X)s y Vλ by the scalar λ. We also make the generalized

eigenspace decomposition g =⊕

gα with respect to ad : g y g. Since g ⊆ gl(V ), we havegα = g ∩ EndK(V )α =

⊕HomK(Vλ, Vλ+α). (Look at the action on the right and on the left,

and track eigenvalues.)

Moreover, ad (σ(X)s) = adσ(Xs), because both act by α on HomK(Vλ, Vλ+α), hence on gα.Hence σ(X)s fixes g, since σ(Xs) does, and so is in n. �

We now turn to a discussion of Cartan subalgebras, aiming for precision. In general, we know thatthey exist and are all conjugate and are nilpotent, at least over C.

Another thing to remember about semisimple algebras — there’s a lot to remember, like there’sno center, that the Killing form is nondegenerate, facts about the derived subalgebra — and here’sanother:

Lemma 31.3: If g is semisimple, h ⊆ g any nilpotent subalgebra, and g =⊕

gα the root spacedecomposition with respect to h. Then the Killing form β pairs gα with g−α nondegenerately∀α. More precisely, β(gα, gβ) = 0 if α+ β 6= 0 **β is a bad choice**.


Let x ∈ gα and y ∈ gβ. For any H ∈ h, (adH − α(H))nx = 0 for some n. So

0 = β((adH − α(h))n x, y) = β(x, (− adH − α(H))n y) (31.1)

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but (− adH − α(H))n is invertible unless β = −α. Nondegeneracy follows from nondegener-acy of β on all of g. �

Corollary 31.3.1: If g is semisimple and h ⊆ h is nilpotent, then the largest nilpotency ideal ing0 of the action ad : g0 y g is 0.


β|g0×g0 is nondegenerate, it’s the trace form of ad, so any ad(g)-nilpotent ideal of g0 must bein radβ = 0. �

We bear in mind the examples of sl and sp, and the claim is that those are typical:

Proposition 31.4: Let K be algebraically closed of characteristic 0. If g is semisimple and h isCartan, then h is abelian and acts semisimply — i.e. h acts diagonalizably.


• By definition of Cartan, h is nilpotent, hence solvable, and by Lie’s theorem we can makeit upper triangular, so h′ the derived subalgebra acts nilpotently, indeed by strictly-upper-triangulars, on any finite-dimensional module, in particular on g. But h = g0, soh′ = 0 by 31.3.1.

• Any X ∈ h that is nilpotent on g must be 0, since KX is an ideal.

• So adXs = (adX)s acts as α(X) in gα. In particular, Xs centralizes h. So Xs ∈ g0 = h,but then X −Xs = Xn ∈ h hence is 0, so X = Xs acts semisimply. �

Corollary 31.4.1: Any subalgebra maximal with respect to these properties is Cartan. Repeating:if g is semisimple, then h is Cartan iff h is maximal diagonalizable abelian.

This gives us another way to see that Cartans exist: 0 is diagonalizable abelian.


We saw that it is abelian and diagonalizable. If h = g0, and h1 ⊇ h is ableian, then h1 ⊆ g0 = h

because it normalizes h. Hence any Cartan is maximal.

For the other direction, given a maximal diagonalizable abelian h, we write g =⊕

gα theweight space decomposition of h y g. We want to show that h = g0, which is the centralizersof h. So if X ∈ g0, then Xs, Xn ∈ g0, and so Xs ∈ h by maximality. So g0 is spanned by h

and ad-nilpotent elements. In particular, g0 is nilpotent (by Engle’s theorem), and thereforsolvable, so g′0 acts nilpotently on g, but it’s an ideal of g0 that acts nilpotently, so g′0 = 0,so g0 is abelian, and now any one-dimensional subspace is an ideal, and a subspace spannedby a nilpotent acts nilpotently, so g0 doesn’t have any nilpotents, so g0 = h. �

Question from the audience: When Cartan subalgebras were originally found, is this how itwas done? Answer: I have no idea. This was all done in the 30s and 40s, and cleaned up in the50s, but I wasn’t born yet. This theory was built up gradually over time, and is too much to thinkup in an afternoon. Certainly, everything that was done was motivated by the classical cases.

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It happens once again that we have five minutes in which to start a new topic. So we do itagain.

The basic idea will be to take a semisimple Lie algebra, pick a Cartan subalgebra. It doesn’tmatter which you pick, because over C they’re all conjugate, and so any combinatorial data youread off doesn’t depend. We want to look at the combinatorial data coming from the root system,axiomatize this a bit, and then show that this data must be one of a small list. Then we need to gobackwards, and build the Lie algebra from the data. The basic ingredient will be the representationtheory of sl2

So, let g be semisimple, h a Cartan — from which we know that the root space is diagonal, notjust generalized eigenspaces —, and g =

⊕α gα, then g0 = h.

Let’s do a quick computation with the Killing form. Let Xα ∈ gα and Y−α ∈ g−α, and H ∈ h.Then

β(H, [Xα, Y−α]) = β([Xα, H]Y−α) (31.2)= −α(H)β(Xα, Y−α) (31.3)

i.e. [Xα, Y−α] = −β(Xα, Y−α)Hα (31.4)

where Hα ∈ h is the unique element such that β(H,Hα) = α(H) ∀H.

In particular, since β is nondegenerate, we can for each nonzeroXα, chose a Y−α so that β(Xα, Y−α) =−1. Thus, we have

[Xα, Y−α] = Hα (31.5)[Hα, Xα] = α(Hα)Xα (31.6)

[Hα, Y−α] = −α(Hα)Y−α (31.7)

If α(Hα) 6= 0, this is an sl2. If it’s 0, then this is a Heisenberg algebra, but this can’t happen,because h acts semisimply on g, but every representation of the Heisenberg algebra acts nilpotently.And the basic idea will be to understand g as a representation of a bunch of sl2s, which will tell usa lot about the structure of the roots.


We pick up where we left off: If g is semisimple over C, h a Cartan, then g =⊕

gα, g0 = h, and ifβ is the Killing form, then β|gα×g−α is a perfect pairing. We take x ∈ gα and y ∈ g−α, and hα ∈ h

such that α(H) = β(H,hα). Then we see that x, y, hα span a Lie subalgebra, which cannot beHeisenberg since hα acts semisimply via ad g. Hence, up to rescaling, 〈x, y, hα〉 is an sl(2). Let’scall it sl(2)α.

We make a few observations. We have that⊕β∈Cαr{0}

gβ ⊕ Chα

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is a subalgebra, on which sl(2)α module. But the weights of any representation are well-behaved:β ∈ Zα/2, and replacing α with α/2 if necessary (if any actual half-integer multiple occurs, thenα/2 occurs), we see that the above sum is over β ∈ Zα. Thus, the subalgebra contains onlyrepresentations V2m, which all have something in weight zero, and so it contains only one irrep,and hence is equal to sl(2)α.

We conclude that ±α are the only non-zero roots in Cα, and that dim g±α = 1. sl(2)α = gα ⊕g−α ⊕ Chα.

A few more facts that we can immediately read off:⋂α 6=0

kerα = Z(g) = 0 (32.1)

∑α 6=0

Chα = g′ ∩ h = h (32.2)

So the αs span h∗ (the dual space) and the hαs span h.

We remark that this picture all works in a reductive algebra, which is a direct sum of a semisimpleand an abelian. Then the Cartan has some extra stuff orthogonal to everything else. In thereductive case, this is called being degenerate.

Proposition 32.1: Let R ⊆ h∗ be the set of non-zero roots α, and R∨ ⊆ h the set of “coroots”hα

def= α∨. Then these satisfy the following properties:

RS1 〈α, β∨〉 ∈ Z for all α and β in R.

RS2 R = −R and R∨ = −R∨, since (−α)∨ = −(α∨)

RS3 〈α, α∨〉 = 2

RS4 If β and α are not proportional, then {〈γ, α∨〉 s.t. γ ∈ (β + Cα)} contains −m, 2 −m, . . . ,m− 2,m for some integer m ≥ 0.

Nondeg R and R∨ span h∗ and h respectively.

Reduced Cα ∩R = {±α}.

These follow from understanding g as a bunch of root spaces. This is given by the decompositionof ad : sl(2)α y g into irreducibles, whence we break g into a bunch of strings, which we can moveup and down by bracketing with xα and y−α.

These are the axioms of a finite root system. **This confused me, because in general we’dneed a way to get between α and α∨. What seems to be going on in “root system” isthat h is a vector space with a nondegenerate pairing 〈, 〉, and R = R∨ is a distinguishedsubset of h, subject to the above conditions. This is the same as saying that we havea vector space h, its dual space h∗, and a chosen isomorphism ∨ : h∗ → h, which sendsR→ R∨. This must be the definition, since often we talk about α ∈ R and then actuallyuse α∨ ∈ R∨.**

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We will classify all finite root systems, and then show that any root system comes from an alge-bra.

But not quite yet: we have a few more corollaries:

We have an explicit formula for the Killing form on the Cartan β|h×h:

β(H,H ′) =∑α∈R

α(H)α(H ′) (32.3)

In particular, β will be invariant with respect to the Weyl group, which we have yet to define, andpositive-definite on hR

def= RR∨.

Moreover, [gα, gβ] = gα+β if α, β, α + β ∈ R. (It’s contained in it, but in fact is all of it.) Thisfollows from the string picture. This will imply that the sl(2)αis for a distinguished subset {αi} of“simple roots” generate, and moreover that g is simple if and only if the root system is indecom-posable.

32.1 Properties of root systems

As we said above, a root system is any set R ⊆ h∗ and R∨ ⊆ h satisfying axioms RS1-4, **andpossibly Nondeg and Reduced?** of 32.1. **Or not? RS4 doesn’t make sense in thislanguage.**

Then RS1 and the fact that R∨ spans h implies that R ⊆ a “real” weight lattice ⊆ h∗: P ={λ s.t. 〈λ, α∨〉 ∈ Z∀α∨ ∈ R∨}. If R spans h∗, we can define Q = ZR the root lattice. ThenQ ⊆ P ⊆ h∗ and both are off full rank, so P : Q is finite. We can thus define the coroot latticeand coweight lattice P ∗ = P∨ and Q∗ = Q∨, and the long and the short is that we can replace theC-vector spaces h and h∗ by real vector spaces hR

def= RQ∨ and h∗Rdef= RQ.

For each α, we define Sα : λ 7→ λ− 〈λ, α∨〉α. This is a reflection fixing the hyperplane 〈−, α∨〉 = 0and sending α 7→ −α. Then Sα y h∗R as above, and thus also hR:

〈λ, Sα(µ∨)〉 = 〈Sα(λ), µ∨〉 = 〈λ, µ∨〉 − 〈λ, α∨〉〈α, µ∨〉 = 〈λ, µ∨ − 〈α, µ∨〉α∨〉

So Sα : µ∨ 7→ µ∨ − 〈α, µ∨〉α∨ is another reflection; Sα = Sα∨ .

We let W be the group generated by the Sαs. This is the Weyl group. And axiom RS4 impliesthat W (R) = R. Question from the audience: Why is W finite? Answer: You can write itdown explicitly as a coxeter group, look at the rank-2 systems, and we will do this next time. Fornow, we remark that W preserves the root-system, and by Nondeg and if R is finite, then W isfinite.

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33.1 Root Systems

Today we work towards the classification of finite root systems. We recall from last time the axioms:A root system is a vector space h, a subset R ⊆ h∗, a subset R∨ ⊆ h, a bijection ∨ : R → R∨, andwe require R be finite, subject to

RS1 〈α, β∨〉 ∈ Z

RS2 R = −R and R∨ = −R∨, with (−α)∨ = −(α∨)

RS3 〈α, α∨〉 = 2

RS4 If α and β are not proportional, then (β + Cα) ∩R is a “string”:

−m•

−m+ 2• . . .

m

•

Nondeg R spans h∗ and R∨ spans h

Reduced Cα ∩R = {±α} for α ∈ R.

Two root systems are isomorphic if there is a linear isomorphism of the underlying vector spaces,inducing an isomorphism on dual spaces, that carries one root system to another.

We define the Weyl group W ⊆ GL(h) generated by the reflections Sα : λ 7→ λ − 〈λ, α∨〉α andhence Sα : µ∨ 7→ µ∨ − 〈α, µ∨〉α∨. Then RS4 implies that W (R) = R, and hence W is finite.

We recall that if R comes from a semisimple Lie algebra, then the Killing form on h is given byβ(H,H ′) =

∑α∈R α(H)α(H ′). This is positive-definite on hR

def= RR. We recall that by RS1 wehave spanning lattices ZR def= Q ⊆ P ⊆ h∗. Then hR is W -invariant.

In any case, there exists a positive-definite W -invariant inner-product (, ) on hR, h∗R: take anypositive-definite inner product, and average it over W . Then Sα acts as an orthogonal matrixw.r.t. this inner product. Hence, being a reflection, Sα(λ) = λ − 2(λ,α)

(α,α) α. So under h∼→ h∗ given

by (, ), we have α∨ ↔ 2α/(α, α). It’s usual in the literature to not distinguish between h andh∗, and work in a Euclidean space, and define it all this way. But that can get confusing, andespecially in the move to infinite-dimensions we don’t have this whole set up. In any case, thisimplies that proportional roots correspond to proportional co-roots, even discarding Reduced.Moreover, (mα)∨ = α∨/m. Each of these has to pair with each other as an integer, and mustpair with themselves to 2: so in the non-reduced case, we can have a system of proportional roots{±2α,±α} and the co-roots are {±(2α)∨,±α∨ = ±2(2α)∨}, so Reduced holds for co-roots if itholds for roots.

Moreover, we see that w(α∨) = (wα)∨ for w ∈W , hence Swα = wSαw−1 for any w ∈W . If V ⊆ h∗

is spanned by a subset of R, then (R,R ∩ V, V ∗, R∨ ∩ V ∗) (where the last one is in quotes) formsanother non-degenerate reduced root system.

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So any α, β not proportional span a rank-2 root system, and we turn to a classification of those.In the process, we will discover a candidate for a new simple Lie algebra. By the way, Reducedimplies that there is only one rank-1 root system, corresponding to sl(2).

So, we have W ⊆ GL(2,R), which is generated by reflections and finite, and there aren’t thatmany finite subgroups of GL(2,R), just cyclic and dihedral groups. But only D is generated byreflections: W ∼→ D2m is dihedral for some m.

But W must preserve Q = ZR, which spans R2. Not many dihedral groups preserve a lattice.If rθ is a rotation by θ, then the eigenvalues are e±iθ and tr(rθ) = 2 cos θ. But to preserve thelattice, W ⊆ GL(2,Z), so the trace must be an integer. So 2 cos θ ∈ {1, 0,−1,−2}. (2 cos θ = 2corresponds to the identity rotation.) Hence θ ∈ {π, 2π/3, π/2, π/3}, i.e. θ = 2π/m for m ∈{2, 3, 4, 6}. These correspond to the rectangular lattice, the square lattice, and the hexagonallattice **twice**:

θ = π θ = 2π/3 θ = π/2 θ = π/3

Then W ∈ {D4, D6, D8, D12}.

A simple observation about dihedral groups: If m is odd, then all reflections in D2m are conjugate.If m is even, then there are two classes, and each generates a Dm. So a generating set of reflectionsmeets every conjugacy class. But the conjugates of Sα are Sβ for other β, so every reflection inW = D2m is Sα for some α. Conversely, if Sα1 and Sα2 generate W , then W preserves the latticeQ = Zα1 + Zα2 spanned by α1 and α2. Moreover, in D2m, we can pick two reflections with amaximally obtuse angle, and these will generate W and the lattice. So we know what the rootsystem is as soon as we know what the lattice is **because two root systems with the sameWeyl group and lattice are related by an isomorphism, by Reduced and Nondeg andthe rest of the axioms**. Thus there are four possibilities:

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m Picture Name notes

2α2 α1

A1 ×A1 = D2 corresponding to the Lie algbera sl(2)× sl(2) = so(4)

3α2 α1

A2 corresponding to sl(3) acting on the traceless diagonals

4

α2α1

B2 = C2so(5) = sp(4). (When we get higher up, the Bs and Cs willseparate, and we will have a new sequence of Ds.)

6

α2α1

G2

a new simple algebra of dimension 14 = number of rootsplus dimension of root space. Smallest rep, we will see, hasdimension 7. We will do this later, when we explain howto construct a Lie algebra from any root system. Thereare lots of ways to describe the Lie algebra. This seven-dimensional representation will come from the Octonians: anon-associative, non-commutative “field”. Then G2 is theautomorphism group of the pure-imaginary part of the Oc-tonians. We will construct explicit matrices for this seven-dimensional rep.

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These pictures are useful. It’s nice to have examples, and moreover we will often refer back to therank-2 case.

These are essentially first-properties of the root systems: we start with the axioms, and haven’tmade any other choices. What have we used? We said it must be dihedral, and dihedral groupsthat preserve a lattice are rare. We never used the full strength of RS4, except to say that Wpreserves the lattice. We could have used a weaker axiom:

RS4’ W (R) = R

and we used that W is finite. There are infinite subgroups of GL(2,Z) generated by reflections.This is a nice correction, because on the face of it RS4 is not symmetric in roots and co-roots, butthis is. We don’t actually want to replace the axiom with this weaker one, because then we’d havethe wrong axioms for an infinite system. But for finite root systems we have

Proposition 33.1: The axioms are symmetric in R↔ R∨.


In the finite case, RS4’ implies RS4. Indeed, all the rank-2 systems that we constructed usingon RS4’ are in fact root systems satisfying RS4. And RS4 only talks about two-dimensionalsubspaces of the root system, so the rank-2 examples are enough. �

Corollary 33.1.1: Each root system has a dual.

Each of the rank-2 ones is self-dual, although not really: dualing switches long with short roots.In fact, we have an example of “Langland’s Duality”: that Bn = so(2n + 1) and Cn = sp(2n) aredual.

Recall how we got here: we started with g a semisimple Lie algebra over C, chose a Cartan h,and got a root system, but the outcome was invariant up to our choice, because all Cartans areconjugate.

Pick a vector v ∈ h∗R such that α(v) 6= 0 for every α ∈ R. We can do this since R is finite and C is

infinite. Then we define the positive root system R+def= {α s.t. α(v) > 0}. Then R− = −R+, and

R = R+ tR−.

You should keep a mental picture, in which we have the root system, an oblique hyperplane, andR+ the part of R on one side of the plane. Then we have a cone R≥0 · R+. Some of the roots inR+ will be interior to the cone, and others not: we define ∆ to be the set of positive roots that areextreme rays of the cone, using Reduced to identify rays with positive roots. Then R+ ⊆ R≥0∆.The αs in ∆ are simple roots, and we will show next time that the simple roots are a basis ofh.

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We consider a finite root system R ⊆ h∗R, and pick v ∈ h such that α(v) 6= 0 for all α ∈ R. Then wedefine R+ = {α s.t. α(v) > 0}, and recall that R = R+tR− where R− = −R+, and R+ generates acone. We let ∆ be the set of extremal rays in this cone; if α ∈ ∆, we call it a simple root. Considertwo simple roots αi, αj ∈ ∆. Then 0 ≥ 〈αi, α∨j 〉 = 2(αi, αj)/(αj , αj), so (αi, αj) ≤ 0.

Lemma 34.1: If v1, . . . , vn ∈ Rn with a positive-definite inner product (, ) satisfy (vi, vj) ≤ 0for all i 6= j, and suppose that there exists v0 ∈ Rn such that (v0, vi) < 0 for all i. Then{v1, . . . , vn} are independent.


Suppose c1v1 + · · ·+ cnvn = 0 with not all ci zero. **missed, but elementary** �

Corollary 34.1.1: Since ∆ generates the cone of R+, hence spans h∗R, we see that ∆ is a basis ofh∗.

Thus, it makes sense to talk about the coefficient of a vector with respect to a simple root.

Corollary 34.1.2: R+ is exactly the set of roots all of whose coefficients are in ∆; every root iseither entirely positive or entirely negative.

Let’s consider R+ r {αi} for αi ∈ ∆. We apply the reflection si = Sαi , and see that si(α) =α− 〈α, α∨i 〉αi ∈ R+ if α 6= αi. Hence R+ r {αi} is fixed by si.

We state the next lemma more generally than needed right now, but we will use it again.

Lemma 34.2: Let ∆ = {α1, . . . , αn} be a set of vectors in Rm with inner product (, ), and assumethat αi are all on one side of a hyperplane: there exists v such that (αi, v) > 0 ∀i. Let Wbe the group generated by reflections Sαi . Let R+ be a subset of R≥0∆ r {0} such thatsi(R+ r {αi}) ⊆ R+ for each i, and such that the set of heights {α, v} s.t. α ∈ R+} is well-ordered (this might follow from the other hypotheses, but we won’t need it to, and we wantit for the induction). Then R+ ⊆W (∆).

In the cases we’re interested in, R+ will be finite, or the set heights will all be positive multiples ofa fixed thing.


Let β ∈ R+. We proceed by induction on its height. Then ∃i such that (αi, β) > 0, becauseif (β, αi) ≤ 0 ∀i, then (β, β) = 0 since β is a positive combination of the αis.

Well, si(β) = β − (positive)αi, and so (v, si(β)) < (v0, β). If β 6= αi, then si(β) ∈ R+ byhypothesis, so by induction si(β) ∈ W (∆), and hence β = si(si(β)) ∈ W (∆). If β = αi, it’salready in W (∆). �

Corollary 34.2.1: For any choice of R+ in a finite root system R, we have R = W (∆), and theSαi for αi ∈ ∆ **generate W**.

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Corollary 34.2.2: R ⊆ Z∆, and R+ ⊆ Z≥0∆.

So knowing just the simple roots determines the root system. We want to know that any choice isthe same.

Proposition 34.3: Any two positive systems R+ and R′+ in R are W -conjugate.

We will again use the fact that the only root that can move out of a root system is αi.


Note: if ∆ ⊆ R′+, then R+ ⊆ R′+. But then R− ⊆ R′− by negating, but these are thecomplements, so R+ ⊇ R′+.

So, suppose αi ∈ ∆ but αi 6∈ R′+. We consider the new system of positive roots si(R′+), whichis again a positive root system, and si(R′+) ∩ R+ ⊇ si(R′+ ∩ R+), because a system of rootsthat does not contain αi doesn’t lose anything under αi. But αi ∈ R′−, so αi ∈ si(R′+) ∩R+,and so |si(R′+) ∩ R+| > |R′+ ∩ R+|. So iterate, and we can’t keep making the intersectionbigger (these are finite sets), so eventually we will find w ∈W such that w(R′+) = R+. �

We’re almost ready to classify root systems, which is the fun part from a combinatorial point ofview. We first define the Cartan matrix:

Let ∆ = {α1, . . . , αn}. We define the Cartan matrix Aij = 〈αj , α∨i 〉. (The columns are the roots,the rows the co-roots. **there should be raised indices**) In terms of the inner product,αij = 2(αi, αj)/(αi, αi). So define di = (αi, αi)/2, and D the diagonal matrix dis on the diagonal.Then (DA)ij = diAij = (αi, αj), so DA is symmetric and positive definite.

Thus A is symmetrizable (there exists invertible diagonal D such that DA is symmetric), andpositive (all principle minors are positive). Indeed, Aii = 2, and Aij ∈ −Z≥0 for i 6= j.

We will start the classification, and tell the answer and sketch how you do it. There are somecalculations to do, checking determinants, and we won’t waste your time calculating determinantson the board.

We introduce the notion of a Dynkin diagram. Any two-by-two minor ofAij will look likeñ

2 −k−l 2

ô,

with kl < 4 by positivity. So either {k, l} = {1,m} for m = 1, 2, 3, or k = l = 0 (if one is 0, theother must be by symmetrizability). So we make a diagram with vertices labeled by (simple) roots.No edge between i and j if k = l = 0. A line if k = l = 1. And an arrow with k edges from i to j

if the i, j block isñ

2 −1−k 2

ôfor k = 2, 3. (The arrow points towards the shorter root.)

If the graph is disconnected, then the Cartan is block diagonal. And it’s clear that any block-diagonal matrix whose blocks satisfy the conditions (symmetrizability and positivity) will satisfythe conditions. So it suffices to classify the indecomposible ones.

Rank 2 We have A2, B2/C2, and G2.

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By symmetrizability, if we have arrows • • •k l

m

, then the third edge must have an arrowof multiplicity m = kl. So k or l is 1, and you can check that the three possibilities all havedeterminant ≤ 0. Moreover, a triple edge cannot attach to an edge, and two double edges cannotattach, again by positivity.

Rank 3 We have A3, B3, and C3.


35.1 Classification of finite-type Cartan matrices

Today we continue with the classification of finite-type Cartan matrices. We recall that to eachfinite root system, by chosing a system of positive roots inside it we get the Cartan matrix Asatisfying Aii = 2, Aij ∈ −Z≥0, which is symmetrizable with positive principle minors. We classifythese by looking at their Dynkin diagrams. We place a node for each positive root, and label theedges based on the two-by-two principle minors.

We said that we need only consider the connected diagrams. At rank 1 there is only A1, and atrank 2 there are A2, B2/C2, and G2. At rank 3, we have only three possibilities: A3, B3, andC3.

In particular, the triple arrow in G2 cannot be extended, and we need never consider it again.There are other diagrams that are forbidden (and hence forbidden as any subdiagram, since anyprinciple minor of A is a valid Cartan; our notion of “subdiagram” is that it is an induced subgraphby removing vertices (you’re not allowed to remove just edges)). For example, a ring of single edgeshas detA = 0, because e.g. giving weight 1 to each vertex we have a vector ~x such that A~x = 0:

•

•

••

•

•

• •· ··

A

( )= 01

1

11

1

1

11

· ··

The rule for weights when looking for such kernel vectors is that an arrow leaving a vertex con-tributes only one neighbor, but an arrow arriving contributes that many neighbors, and A is singularif there is a weighting such that each vertex has twice as many neighbors as its own weight. Here

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are some more principle families of forbidden diagrams:

• • · · · •

•

•

•

•

•

•

•

•

•

• • · · · •

•

•

•

• • · · · •

•

•

•

• • · · · •• •

• • · · · •• •

• • · · · •• •

Here are some more forbidden subgraphs:

• • • • • det = 01 2 3 2 1

• • • • • det = 02 4 3 2 1

• • det = −4

• •

••

•

•

•det = −5

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• • det = −8

• •

• • not symmetrizable

• •

So we can never have more than one double edge, more than one trivalent vertex, both a trivalentand a double edge, or any circuits. Hence any diagram with a double edge is a chain. The doubleedge must come at the end, or can have only one edge on each side:

• • · · · • •Bn

• • · · · • •Cn

• • • •F4

If there are no double edges, we can have the chain An: • • · · · • . If there is a branching,one can check that the determinant of a branching where the arms have lengths k, l,m (including themiddle vertex), then the determinant is klm( 1

k + 1l + 1

m−1). So this is only positive if 1k + 1

l + 1m > 1.

So it’s forbidden to have three arms of length three or more, since 1 = 13 + 1

3 + 13 , and it’s forbidden

to have two arms of length four or more, since 1 = 12 + 1

4 + 14 .

321

2 1

2 1 4

3 2

2

1

3 2 1

These allowDn, and the only other possibility is a branch with one arm of length 2, and one of length3. But there is one more Egyptian-fraction way to add three fractions to get 1: 1 = 1

2 + 13 + 1

6 .

6

3

42 5 4 3 2 1

Thus, the E family only has three entries: E6, E7, and E8.

•• • •

• •

•• • •

• • •

•• • •

• • • •

E6 E7 E8

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Now we will just observe that the infinite families are groups we already know: sln+1 An,sp2n Cn, so2n+1 Bn, and so2n Dn. The two most closely related groups are Bn and Cn.Indeed, they have dual root systems, in the sense that their Cartan matrices are transpose of eachother, and this leads to the Langlands program.

It would be nice to see where E6, E7, E8, F4, and G2 come from. This is somehow redundant,because we could just present them, and we will go quickly.

To get from a Cartan matrix to a root system, we observe that ifñd1

d2

ô ñ2 −l−k 2

ôis sym-

metric, then d1l = d2k, so d1/d2 = l/k > 0. So A can be symmetrized with a positive D, and DAis positive-definite. Thus, we can find vectors αi so that diAij = (αi, αj , and di = (αi, αi)/2. Thus

Aij = 2(αi, αj)/(αi, αi)def= 〈αj , α∨i 〉. So we can start constructing out root system. And we can let

W be generated by Sαi : this leaves (, ) and the lattice Q = Z{α1, . . . , αn} invariant. Then we letR = W · {α1, . . . , αn}. Then all the axioms of a root system are trivially satisfied (the root-stringaxiom follows just from having the root system fixed by W ), except for the fact that it may benon-reduced. Indeed, if we started with a Cartan matrix not of finite type, we could get infiniteroot systems that are not reduced.

But the reduced-ness of the finite-type ones comes from the classification: For An, Dn, and En, thematrices are already symmetric, di = 1, and the roots are all the same length. If there is a doubleor triple arrow, there’s only one, meaning that there are only two lengths of roots, and di ∈ {2, 1/2}or di ∈ {3, 1/3}, and so the systems are reduced.

By the way, the simplest non-reduced system comes from • ⇒ • ⇒ •.

Let’s compute an example, and figure out how big is F4 = • − • ⇒ • − •. We have four simpleroots. If we start at α1 = 〈1000〉 and start reflecting, we get 〈1100〉, 〈1120〉, 〈1220〉, 〈1122〉, 〈1222〉,〈1242〉, 〈1342〉, 〈2343〉, and this is as high as we can get. These are all the vectors we can getkeeping positive coefficient with α1. Starting with α2 = 〈0100〉, we get the root system of typeC: 〈0120〉, 〈0122〉, 〈0011〉, 〈0110〉, 〈0111〉, 〈0121〉. And that should be all the positive roots: foursimples, and fourteen (?) more. The point is that, at least in principle, just from reading thediagram we can compute the root system. Then there are 36 total roots, and so if F4 comes froma simple Lie algebra, which it does, then dim f4 = 36 + 4 = 40 (for the four-dimensional Cartansubalgebra).

In fact, it turns out that for any linear diagram, W is the symmetry group of a regular polyhedronin 4-space. • = • is the symmetry group of a square, and • − • is a triangle. **missed the endof the geometry**

There is something more general, called Coxeter groups, which preserve a polyhedron but nota lattice. These allow the pentagonal edge. There are the infinite families (polygons, simplex,cube/octahedron), and some sporadic ones with pentagons only in dimensions 4 and less.

The Es are also interesting. Let’s talk about E8, where the seven vertices along the bottomare ei − ei+1 for i ∈ {1, . . . , 7}, and the ei are orthogonal in R8. Then the one sticking up is

106

−12(e1 + e2 + e3) + 1

2(e4 + · · · + e8). Then the Weyl group is generated by these eight reflections,and using a computer one can construct the whole root system.


A new problem set is available. Many of the exercises are of the form “pick your favorite rootsystem and verify X”. These are part of a beautiful subject concerning the theory of Weyl groups:invariant polynomials, exponents, etc. Most of this holds for Coxeter groups more generally.

36.1 From Cartan matrix to Lie algebra

Today we begin the process of reversing the map we made so far: from simple Lie algebra to rootsystem to Cartan matrix to the classification of root systems of finite type. We want to now seethat all these finite roots systems are the root systems of simple Lie algebras, and that a Lie algebrais determined by its Cartan matrix.

Here’s the philosophy. We take e.g. the root system of G2 and pick a line giving a positive sys-tem.

α2α1

Then there are two extremal hence simple roots, so the Cartan h should have dimension 2. Andthen the elements of the Lie algebra are in the picture given by the root strings: for each line,there’s an ei, an fi, and an hi = [ei, fi]. So what we should try to do is just take the algebragenerated by these strings and relations. It turns out that this won’t have enough relations, butwe’ll show that what we get has a unique maximal ideal, which we will mod out by to get a simpleLie algebra.

Most of the construction will go through knowing that the Cartan has 2s on the diagonal, non-positive off-diagonals, and if Aij = 0, then Aji = 0. These “generalized Cartan matrices” give“Kac-Moody algberas”, which we won’t talk about — we will eventually use that these matricesare of finite type — but Kolya will probably talk about them.

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So, define g = 〈ei, fi, hi : i ∈ {1, . . . , n}〉, with the relations that

[hi, ej ] = 〈αj , α∨i 〉ej[hi, fj ] = −〈αj , α∨i 〉fj[ei, fj ] = δijhi

[hi, hj ] = 0

I.e. we have n sl(2)is, with h = 〈h1, . . . , hn〉 acting by weights, and [ei, fj ] = 0 if i 6= j.

This is a quotient of the free Lie algebra, of course. We can Q-grade (Q is the root lattice) the freeLie algebra F (ei, fi, hi : i = 1, . . . , n) by giving deg ei = αi, deg fi = −αi, and deg hi = 0. Then therelations are homogeneous, so g inherits the same grading. By construction, ad h acts diagonallyby the grading, because we included it in the relations.

We now define a triangular decomposition. Let n+ be the subalgebra generated by the eis, n− bythe fis, and h = 〈h1, . . . , hn〉. This is called the “triangular decomposition”, because n+ are thestrict upper-triangular ones, n− the lower-triangulars, and h the diagonal matrices.

Proposition 36.1: g = n− ⊕ h⊕ n+


It’s enough to show that g = n− ⊕ h⊕ n+, because they have nothing in common (⊕ followsby grading). Well, (ad fi)n− ⊆ n−, (ad fi)h ⊆ 〈fi〉 ⊆ n−, and (ad fi)n+ ⊆ h + n+. So ad fipreserves n− + h + n+, and also h does obviously (it acts diagonally), and by symmetry n+

also preserves the sum. So n− + h + n+ is an ideal of g and in particular a subalgebra, but itcontains all the generators. �

It’s not obvious that everything we’ve done doesn’t just collapse. To show that e.g. the his remainlinearly independent, we need a representation. We define the Verma module Mλ or g for any fixedlinear functional λ ∈ h∗ (a “weight”), where h is the space with basis {α∨i = hi}.

Proposition 36.2: There exists an action of g on C〈f1, . . . , fn〉 · vλ **vλ is just a symbol**such that

fiÄ∏

fjkvλä

=Äfi∏

fjkävλ

hiÄ∏

fjkvλä

=

(λ(hi)−

∑k

〈αjk , α∨i 〉)·Ä∏

fjkvλä

eiÄ∏

fjkvλä

=∑k:jk=i

fj1 . . . fjk−1fjk+1

. . . fjl vλ


We just have to check the commutators. The h relations are OK by degree considerations.The other relations are [ei, fj ]. By construction, if i 6= j, then ei can’t see the fj tacked on athe beginning. What about [ei, fi]? eifi

Äfvλä

= hifvλ + fieifvλ, clear by construction. �

108

This justifies the relations. We have an explicit module on which g acts. Because λ is arbitrary,it’s a corollary that the linear space spanned by the his does not collapse when we mod out: in thisrepresentation, hi acts with eigenvalue independent of the other hjs.

Corollary 36.2.1: h = h, i.e. h ↪→ g, and also ei and fi do not map to 0 in g. Indeed, the fis justact freely. So n− and n+ are free on {fi} and {ei}. But we won’t use this last fact.

We haven’t used much of anything about the Cartan matrix A. Let’s now assume that A isnonsingular and indecomposable, meaning that the Dynkin diagram is connected. (Think of thediagram as a directed graph: we have a directed edge for each non-zero entry in the matrix. We’reassuming that the graph is strongly connected: you can get from anywhere to anywhere else byfollowing arrows.)

Proposition 36.3: Every proper ideal of g is homogeneous and contained in n−+ n+, and doesn’tcontain any ei or fi.


The adjoint action of h gives the grading, and so any ideal is homogenous. The details arean exercise: the sum of elements of different degrees, by bracketing with h gives differentweights, and you do this enough to get an invertible matrix, from which you can pick out thehomogeneous elements themselves.

Suppose that we have an ideal a containing H ∈ h with H 6= 0. Then there exists i such thatαi(H) 6= 0, since A is nonsingular. (If A is singular, then ker ad h is in the center of g, andis a proper ideal. But then mod out by it and get an algebra that behaves as above.) Then[H, ei] = (6= 0)ei, so ei, fi ∈ a, i.e. a contains an 〈ei, fi, hi〉. And if we contain an ei or an fi,then again we contain an sl2. Thus the ideal is not proper by strong-connectedness. �

Corollary 36.3.1: g has a unique maximal proper ideal, i.e. a unique quotient g � g, and h +〈ei, fi〉 ↪→ g.

Up to this point we used essentially nothing about A. 2s on the diagonal, nonsingular didn’t reallymatter because we could have started by modding out by the center, and indecomposable. Tounderstand g, however, we need some serious information about A.

Proposition 36.4: Serre Relations

The relations

(ad ej)1−〈αi,α∨j 〉ei = 0 (36.1)

(ad fj)1−〈αi,α∨j 〉fi = 0 (36.2)

hold in g.


By symmetry, it’s enough to do the f relations. Let S be the left-hand-side of (36.2). It’senough to check the bracket with e: we claim that [ek, S] = 0 in g for any k. The cases when

109

k 6∈ {i, j} are trivial. But ad ej kills fi, and adhj(fj) = mfi, where m = −〈αi, α∨j 〉. So thepicture is that we have an sl(2)i, and fi ∈ g is acting like a highest-weight vector. We get aninfinite string {fi, (ad fj)fi, (ad fj)2fi/2, . . . }, which is an infinite sl(2)j-submodule of g.

But we know that at the end of the day the string is finitely long, with length m. This is theSerre relator: (ad ej)(adfj)m+1fi = 0.

That [ei, S] = 0 follows similarly. We have not used the full symmetrizability of A, but wedo use that if one off-diagonal is 0, so is its transpose.

We haven’t quite completed the proof. We will pick it up next time. �


**I missed the first half of the class. We completed the proof of the Serre relations,which involved various lemmas.**

Proposition 37.1: Serre relations.

The Serre relations force the property that the integrable **defined when I wasn’t here**elements of g form a submodule.

Corollary 37.1.1: (g, ad) = I(g)

What this means is that the adjoint action of g on itself for each i breaks into finite-dimensionalsl(2)-strings. This lets us almost get the finite-dimensionality, except we haven’t used that A isfinite-type. Everything we’ve done so far uses only that Aii = 2 for each i, and that Aij ∈ Z≤0 withAij = 0 iff Aji = 0. So if we don’t assume that A is finite-type, then we get infinite-dimensionalKac-Moody algebras that are integregrable with respect to themselves.

Corollary 37.1.2: The non-zero weights of g (a subset of Q = Z〈α1, . . . , αn〉) satisfy the axioms ofa root system R with R+ = {weights of n+}, except that the weight system is not necessaryfinite and not necessarily reduced.

In particular, it will be invariant under the Weyl group generated by the reflections across theαis. Now we assume that A is finite-type. Then 〈αi, α∨j 〉 = (αi,

2αj(αi,αj)

) for some positive-definitesymmetric (, ) on Rn with a basis α1, . . . , αn.

Question from the audience: Remind me what finite-type means? Answer: It means on theABCDEFG list: symmetrizable and with positive principle minors.

Then the si being the reflections in αi generate a finite group W .

We recall a technical lemma:

Lemma 37.2: If R+ ⊆ Rn has the properties that

(a) 0 6∈ R+

110

(b) there exists a v such that (R+, v) > 0 and that this set is well-ordered

(c) si(R+ r {αi}) ⊆ R+

then R+ ⊆ W (∆). (Where ∆ is the set of αis, which give the reflections si, and Rn has aninner product.)

Then the non-zero weights of g are finite and reduced, since R+ satisfied the conditions to thelemma.

Corollary 37.2.1: If A is finite-type, then R is a finite reduced root system with Cartan matrixA.

But A determines R and R determines g, and then g is a finite-dimensional simple Lie algebra withroot system associated to A.

Even more, g is definable over Q. **Did we use anything about characteristic?**

We finish with showing that g is the unique Lie algebra with the root system giving A. First,any Lie algebra with a simple root system is simple: it has a highest element, which generates thewhole thing, and any linear combination of roots generates the highest root. So suppose that g1

is another simple Lie algebra with the same root system as g. Then all the relations we used todefine g hold in g1, because we can repeat the proof, finding sl(2)s, and by definition the weightsact appropriately, etc. Thus we have a homomorphism g � g1. But g1 is simple, so the kernelmust be the unique maximal proper ideal of g. So the root system determines the algebra up toisomorphism.

In the homework, you work out the root systems of the orthogonal groups. The degenerate rootsystems of type D are the same as other root systems: so(6,C) ∼= sl(4,C) and so(4,C) ∼= sl(2,C)×sl(2,C).

Moreover, we have five Lie algebras that are not sl, so, or sp. By the way, there’s that homeworkthat’s up, and soon there will be a few more problems.

There’s more. Next time, we will begin determining the representation of these Lie algebras, whichalso follows from the combinatorics.


To fully understand the classical Lie algebras, we want to know their representation theory. Itsuffices to understand the irreducible representations, as these algebras are semisimple.

Recall the classification of the irreps of sl(2). We have generators e, f , and h. We know that h actsdiagonally, and we think of f as moving us up and e down the string. If h acts on the top vectorv0 by m, then we did a calculation that shows that efm+1 = 0. So everything below fm+1v0 is asubmodule; modding out by it gives an irreducible.

111

This will be the general method: we will construct a module, and then mod out by its maximalproper submodule. For sl(2), we can know easily that there must be a maximal proper (possiblyzero) submodule of the module generated by any given vector of a given weight.

38.1 Irreducible modules over g

From last time, we have g decomposed as a vector space as g = n−⊕h⊕n+. By the PBW theorem,we can pick monomial bases in this order. As a vector space:

U(g) ∼= U(n−)⊗ U(h)⊗ U(n+)

Fix λ ∈ h∗ a for-now-arbitrary linear functional. Let’s look at b = h⊕n+. Then n+ is an ideal of b,and so b has a one-dimensional module C · vλ, where the action is that for H ∈ h, Hvλ = λ(H) vλ,and n+vλ = 0.

From this we define the Verma module Mλ = U(g)⊗U(b) C · vλ. As a vector space this is U(n−)⊗CC · vλ, and Mλ is presented by the generator vλ and relations Hvλ = λ(H) vλ and n+vλ = 0 (norelations on how n− acts).

Mλ has a weight grading. We let Q+def= N{α1, . . . , αn} ⊆ Q

def= Z{α1, . . . , αn}, and then

Mλ =⊕β∈Q+

(Mλ)λ−β (38.1)

Every proper submodule N ⊆ Mλ has N ⊆ ⊕β∈Q+r{0} (Mλ)λ−β (since (Mλ)λ generates), and sothere is a unique maximal proper (possibly zero) submodule — the sum of all the proper submodules— and an irreducible quotient Mλ � Lλ. Question from the audience: The quotient is graded?So the submodule is graded? Answer: All submodules are graded by the action of h.

This Lλ is irreducible, but possibly infinite-dimensional. We saw last time that every g-module hasa submodule of integrable elements. Well, Lλ is irreducible, so either every element is integrable ornone are. We define P to be the linear functions that are integrable **integral?** P def= {λ ∈ h∗ :〈λ,Q∨〉 ⊆ Z}, and define the dominant integral weights to be P+ = {λ ∈ P : 〈λ, α∨i ≥ 0 ∀i}. Thedistinction P v.s. Q will matter for the classification of Lie groups. We have h ⊇ P∨ ⊇ Q∨ andh∗ ⊇ P ⊇ Q.

Proposition 38.1: If λ ∈ P+ then Lλ is integrable.


Enought to check vλ is integrable. We know that eivλ = 0, and hivλ = 〈λ, α∨i 〉vλ, where〈λ, α∨i 〉 = m ≥ 0 is an integer. By the sl(2) picture, we know that eifm+1

i vλ = 0, and indeedejf

m+1i vλ = 0 for every j. So in Mλ, fm+1

i vλ generates a submodule, so is 0 in Lλ. �

We write the additive group h∗ multiplicatively: λ 7→ xλ. This is formal notation: xλxµ = xλ+µ.We do this so that we can write the group-algebra Z · h∗, which are “polynomials”

∑λi cix

λi . Wecan also consider “formal power series”, which form a completion of this module.

112

We consider M as a g-module with diagonal h-action and finite-dimensional weight spaces, andsuch that there exists a finite set S such that the weights of M are in S + (−Q+). Then we saythat M is in the category O **a full subcategory of g−Mod?**. The standard thing is towork with the category O, which imposes another technical condition, but is awkward to work withand totally unnecessary. We write ch(M) def=

∑λ a weight of M dim(M)λ · xλ. This is a finite linear

combination of power series in x−αi times monomials xλ. It’s a purely formal way to keep track ofthings; ch(M) is some sort of fraction formal Laurant series.

Example: ch(Mλ) =xλ∏

α∈R+(1− x−α)

. We understand x−α as a product of x−αi for α =∑αi,

and the division as multiplication by the formal geometric series.

Then it’s clear that O is closed under submodules, quotients, extensions, tensor products.

Corollary 38.1.1: **We assume that λ ∈ P+.**

(a) ch(Lλ) is W -invariant. “I just said what the proof is, so I don’t need to write it down.”

(b) If µ is a weight of Lλ, then µ ∈W (ν) for some ν ∈ P+∩ (λ−Q+). (By the way, λ ∈ P+,so let’s say ν ≤ λ if λ, ν ∈ P+ and λ− ν ∈ Q+.)

We implicitly proved this already. P = W (P+), since if λ ∈ P r P+, then 〈λ, α∨i 〉 < 0for some i, so si(λ) = λ + (positive integer)αi, and so we can move things higher. ButW is finite, so W (λ) has a maximal-height element, which must be in P+. This showssomething more: if λ ∈ P+, then W (λ) ⊆ λ−Q+. This implies (ii).

(c) Lλ is finite-dimensional.

This follows from (b). The Weyl group is finite-dimensional. R+P+ and −R+Q+ aretwo cones pointing opposite direction, since the inner product is evaluation of a positive-definite form. The intersection of the two cones is 0, since the inner product of anythingin R+P+ with anything in −R+Q+ if ≤ 0. So we can find a hyperplane separating thecones: there’s a linear functional η positive on P+ and negative on Q+. Then λ − Q+

is below the η = η(λ) plane. But −Q+ is spanned by these −αis, each of which has anegative value under η, so we can only subtract finitely many from λ and keep η ≥ 0.So P+ ∩ (λ−Q+) is finite.

(d) Every finite-dimensional irreducible g-module is Lλ for a unique λ ∈ P+.

Because we pick a vector, move up by eis until we can’t go up any more and get a topweight, and by the sl(2) this top weight is in P+. By irreducibility, this top weight vectorgenerates, and we send vλ to it to get a map Mλ � our module. But Mλ has a uniquemaximal submodule, and since our module is irreducible, the map must be modding outby that module.

This is good, but doesn’t give us particularly good descriptions of the details of the irreps.

We now define a special element ρ ∈ P+, by 〈ρ, α∨i 〉 = 1. Then ρ =∑i Λi, where by definition

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〈Λi, α∨j 〉def= δij **In Problem Set 5, we will call Λi a fundamental weight.**. Notice that

P+ = N{Λ1, . . . ,Λn}. We let ρ = 12

∑α∈R+

α. Then we know that si(R+r {αi}) = R+r {αi}, andsi(αi) = −αi. So si(ρ) = ρ− αi, hence 〈ρ, α∨i 〉 = 1, so ρ = ρ.

We will prove the following next time:

Theorem 38.2: We let ε : W → {±1} be the sign function si 7→ −1, i.e. w 7→ det(w). Then thecharacter of λ is defined as χλ def= ch(Lλ). It is given by the following formula:

χλdef=

∑w∈W

ε(w)xw(λ+ρ)−ρ∏α∈R+

(1− x−α)(38.2)

A priori this could be an infinite sum, but in fact xρ∏α∈R+

(1− x−α) =∏α∈R+

(xα − x−α). Soboth the numerator

∑w∈W ε(w)xw(λ+ρ) and denominator xρ

∏α∈R+

(1− x−α) are antisymmetricwith respect to W , so the whole expression is W -invariant. So this is an expansion of a rationalfunction but it’s also W -invariant, so it must be a polynomial.


**I was out of town.**

Lecture 40 December 1, 2008

Today we will do some algebraic geometry, but first we study sl(2,C). This is Lie(SL(2,C)), whichis an algebraic group, but it has a non-trivial center: Z(SL(2,C)) = {±1}. We can mod out bythis center and get PSL(2,C) def= SL(2,C)/{±1} = GL(2,C)/{scalars} def= PGL(2,C).

SL(2,C) is simply connected. The easiest way to see this is to take the upper-triangular ma-

trices B = {ñ∗ ∗0 ∗

ô∈ SL(2,C)}. Then SL(2,C) y C2, hence acts on the lines in C2, i.e.

SL(2,C) y P1(C) = S2. A good parameterization of S2 is the equivalence classes 〈(1, z)〉 and〈0, 1〉. It’s clear that B fixes the point 〈1, 0〉 ∈ P1, and indeed is the stabilizer of this in P1, soP2 = SL(2,C)/B.

This isn’t what we want to do. Let’s use U = {ñ

1 ∗0 1

ô∈ SL(2,C)}, which is the stabilizer of the

vector (1, 0) ∈ C2r{0}, on which SL(2,C) acts transitively. So SL(2,C)/U ∼= C2r{0} ∼= R4r{0}as real manifolds. But U ∼= C, so SL(2,C) is simply connected.

Of course, PSL(2,C) is the adjoint group: we’ve modded out by everything we can. Hence SL(2,C)and PSL(2,C) are the only connected groups with Lie algebra sl(2,C).

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Let’s study how these groups interact with the representation theory. Remember that sl(2,C)is generated by {e, f, h}, and in any representation, e moves up the chain, f down, and h actsdiagonally with eigenvalues changing by 2 from m at the top to −m at the bottom:

•v0

•v1

•v2

...

•vn−1

•vn

h=myy

h=m−2zz

h=m−4zz

{{

h=2−mzz

h=−mzz

F=1��

m=E

HH

F=2��

m−1=E

HH

HH

��

JJ

F=m��

1=E

HH

Well, we have

SL(2,C)

sl(2,C)

ker=2πiZhexp

::uuuuuuuuuuuu

ker=πiZhexp

$$IIIIIIIIIIIIth =

ñt−t

ô7→ñet

e−t

ôPSL(2,C)

So PSL(2,C) acts on a representation Vm only with m even, because −1 ∈ SL(2,C) acts on Vm as(−1)m. m is really λ ∈ P , and m is even only if m ∈ Q.

This will be the general picture.

simple g/C Vλ λ ∈ P+

simply connected Gs y Vλ Gs → GL(Vλ)

**well, this seems to be less a table and more boardwork that vaguely lines up incolumns**

Consider the Cartan h ⊆ g. Then exp : h � T ⊆ G, where T = h/lattice. If G acts faithfully onVλ**lost** Well, exp(2πiα∨j ) is trivial on every Vλ.

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On the other hand, for each α ∈ R, we have sl(2)α ⊆ g, spanned by Xα, X−α, and hα = α∨. Thiswill break a representation V into root strings. It might happen that every string has m even, inwhich case this action will factor through PSL(2).

So the only cases are that α∨/2 acts trivially, or else α∨ generates the kernel of the action of thissl(2).

**Boardwork is erratic, and I’m a little lost.** The moral of the story is that the lattice in“T = h/lattice” contains 2πiQ∨ and is contained in 2πiP∨. “Maybe I don’t quite know this yet.”At least we can say that the lattice is contained in πiQ∨.

The other way to see this, which is the right way: the biggest this lattice can be is when I mod outby the center of the group, which leaves the adjoint action. So “lattice” is contained in the latticeof things that act trivially in the adjoint action Vθ = (g, ad). This has weights spanning Q. So weknow that any element of h that is integral on every root, i.e. anything in P∨ — 2πi times it willexponentiate to something that acts trivially in the adjoint group.

We didn’t plan to say all this, but rather give a flavor. But the picture should basically be this: Youhave some Lie group and Lie algebra. You look at the Cartan and its kernel under the exponentialmap. In the simply-connected group, this kernel is 2πQ∨, and in the adjoint group, the kernel is2πiP∨. For any given Vλ, the kernel will be between these two, and our claim is that T = exp(h)contains the center of the simply-connected group.

This last part is a little hard, and uses some algebraic geometry. What we will do is construct analgebraic group with the right Lie algebra, that acts faithfully on the Vλs, and we will see that itis simply-connected.

The model is SL(2)/PSL(2), but we will mention one more example. We will see that SL(n,C) issimply-connected, and of course Z(PSL(n,C)) = nth roots of unity. So we can mod out by anysubgroup of this cyclic group of order n, but at the bottom we mod out be all of them and getPSL(n,C) = SL(n,C)/Z. Since this is type-A, we can identify roots and co-roots: αi = εi − εi+1.We think of h∗ ⊆ Cn as the space of vectors the sum of whose coordinates is 0; this is of course(n− 1)-dimensional.

Let’s say this again. h is the diagonal traceless matrices: h = {〈z1, . . . , zn〉 ∈ Cn :∑zi = 0}.

So h∨ = Cn/C〈1, . . . , 1〉. α∨i = εi − εi+1, and the dual basis is λi = 〈1, . . . , 1, 0, . . . , 0〉 with atleast one 0. So basically P∨ = Zn/〈1, . . . , 1〉, and Q∨ is the sublattice of P∨ spanned by the λi.So P∨/Q∨ = Z/nZ, because /Q∨ makes all the coordinates equal, and P∨ mods out when thisnumber is n.

40.1 A little about algebraic geometry

We want to produce from g an algebraic group. In fact, we will use the representation theory,and we will want to produce actually a bunch of groups, modding out by different lattices. So wewill work with a lattice between P∨ and Q∨, and work with the category of representation all of

116

whose weights are in this lattice. This subcategory of the category of all representations will stillbe monoidal abelian. We will need to build the algebraic functions from the group to the matrixentries of the action out of the V s.

We will only be concerned with affine varieties over C, meaning that we have the locus of a systemof polynomials in Cn: X = {~x ∈ Cn : pi(~x) = 0∀i}, for some chosen polynomials pi ∈ C[x1, . . . , xn].Consider I the ideal in C[~x] consisting of all functions that vanish on the variety. Then pi ∈ I, butin general I will not be generated by the pis; rather, I is the radical of the ideal generated by thepis.

We define O(X) def= C[~x]/I, thinking of it as “the algebra of polynomial functions on X”. Indeed,we have a well-defined map from O(X) → C(X), and in fact O(X) is the ring of functions onX ⊆ Cn generated by the coordinate functions on X. Then O(X) is a finitely-generated C-algebra(in fact it’s reduced, i.e. the only nilpotetns are 0, but we will ignore this issue, citing generalfacts).

What would it mean for X to be a group with algebraic multiplication? The group law X×X → Xturns into a comultiplication ∆ : O(X) → O(X) ⊗ O(X). This is a general rule: a morphism ofalgebraic varieties is just a homomorphism of the algebra of functions.

The game will be to see what kind of map ∆ makes X into a group, and then to build this algebraicstructure.


41.1 Review of Algebraic Geometry

Let X ⊆ Cn be an affine variety: X = V (P ) = {~a s.t. p(~a) = 0 ∀p ∈ P} is the vanishing set of aset P ⊆ C[x1, . . . , xn]. We can associate to X its ideal I = I(X) = {p ∈ C[~x] s.t. p|X = 0}; this isa radical ideal, and P ⊆ I implies V (I) ⊆ V (P ). We define the polynomial functions on X to bethe ring O(X) = C[~x]/I(X).

A morphism of affine varieties is a function f : X → Y such that the coordinates on Y of f(x)are polynomials in the coordinates of X; this gives a function by precomposition f# : Fun(Y ) →Fun(X), and is a morphism if f# : O(Y )→ O(X) is a homomorphism of C-algebras.

Any point a ∈ Cn gives an evaluation map eva(p) = p(a) : C[~x] → C. Then a ∈ X iff I ⊆ ker evaiff eva : O(X) → C. Going the other way, O(X) determines X as the set of evaluation mapsO(X)→ C. So morphisms of affine varieties are exactly morphisms of their algebras of functions,going the other way. Algebraic Geometry is all about driving backwards: you go along the highway,looking out the rear-window the whole time.

We remark that if X ⊆ Cm and Y ⊆ Cl are affine varieties, then X × Y ⊆ Cm+l is an affinevariety. What’s its coordinate ring? The projections X × Y → X,Y gives maps O(X),O(Y ) →O(X × Y ); thus O(X) ⊗C O(Y ) → O(X × Y ). But in fact this is an isomorphism, because

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everything is finitely generated by the coordinate functions and evaluation of functions at pointsseparates functions.

As in the very first lecture, we define an affine algebraic group to be a group in the category ofaffine algebraic varieties. Let’s unpack the multiplication µ : G×G→ G, inverses i : G→ G, andidentity e : {pt} → G. We get

• A comultiplication ∆ : O(G)→ O(G)⊗C O(G), which must be coassociative.

• An antipode S : O(G)→ O(G)

• An evaluation ε = eve : O(G)→ C

We get the axioms of a commutative Hopf algebra. We saw before that the universal envelopingalgebra of a Lie algebra is a cocommutative but not commutative Hopf algebra; O(G) will becommutative but not generally cocommutative. U(g) and O(G) are dual in a sense that we will tryto make precise.

First, a remark. We never required that O(G) be reduced. A non-reduced Hopf algebra is a groupscheme. But in fact it cannot happen that a finitely-generated commutative Hopf algebra over Cbe non-reduced. Also, G is smooth.

We toss out results from algebraic geometry without justification. A reduced scheme has anonempty set of smooth points, and the group law acts transitively. A non-reduced scheme mighthave no smooth points, but we cannot have a non-reduced group in characteristic 0. We say that avariety X over C is smooth if X is a manifold; there is an algebraic definition which we skip.

Let’s explain why we cannot have a nonreduced group. Let m = ker ε; the fact that e · e = e meansthat

O(G) ∆ //

ε

��

O(G)⊗O(G)

ε⊗ε��

C C⊗ C∼oo

(41.1)

commutes. Hence∆(m) ⊆ m⊗O(G) +O(G)⊗m. (41.2)

This makes m a Hopf ideal; (41.2) is exactly the condition needed to make O(G)/m a Hopf al-gebra. Hence ∆(mn) ⊆ ∑

k+l=n mk ⊗ ml, and so ∆ and S induce ∆ and S on R = grmO(G) def=⊕k∈N mk/mk+1. R also has an evaluation by modding out by m, so R is a graded Hopf algebra. R

is generated by R1, and if x ∈ Ri, then each x is primitive:

∆x = x⊗ 1 + 1⊗ x (41.3)

Well, R = C[y1, . . . , yn]/J , where n = dimG and J is a Hopf ideal in the Hopf algebra C[~y] wherethe yis are primitive. Well, C[~y]⊗C[~y] = C[~y, ~z], and ∆ : f(~y) 7→ f(~y+ ~z). What are the minimal-degree homogeneous elements of J? (We can assume that J1 = 0 wolog, because we can take the

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yis to be a basis of R1.) It must be primitive **why?**, which means that f(~y+~z) = f(~y)+f(~z).In characteristic 0, this forces f to be homogeneous of degree 1

Question from the audience: What about the antipode? Answer: Well, it doesn’t matterfor us: we’ve shown that a bialgebra must be smooth. But we can do a similar calculation: if∆(f) =

∑f1 ⊗ f2, then we must have

∑f1 ⊗ S(f2) = ε(f). But if f is primitive, then we see that

fS(1) + 1S(f) = 0, and so S(f) = −f is smooth.

In any case, the algebraic definition of a smooth whatsit is that G is smooth at e if grmO(G) is apolynomial ring. We leave out why this agrees with the earlier definition.

Anyway, as an algebraic variety G has the Zariski topology, but as an analytic variety it sits inCn, and in particular is a Lie group. We’d like to understand how to get the Lie algebra of ouralgebraic group algebraically, and eventually how to go the other way. Everything we’ve done upto now makes sense over any algebraically closed field, although we needed characteristic zero forsmoothness.

41.2 Algebraic Lie algebras

We can think of Lie(G) as the tangent space at the origin, but this doesn’t give us the algebrastructure. Better is to think of g = Lie(G) as the left-invariant vector fields, and then U(g) acts asleft-invariant differential operators on S(G) the smooth (analytic) functions on G. It’s not obviousthat this action takes polynomials to polynomials.

Let U be a left-invariant operator on S ⊆ Fun(G), with S a left-invariant algebra containing O(G).For any function, we can make sense of the coproduct, by composing with the group product. Thespecial property of algebraic functions is that ∆(f) is a finite sum

∑f1⊗f2. If f is a polynomial, we

use bad indices, and fi are characterized by f(xy) =∑f1(x)f2(y) for x, y ∈ G. Well, if f ∈ O(G)

and g ∈ G, we define the action Gy O(G) by gf = f ◦ g−1, i.e.

(gf)(h) = f(g−1h) =∑

f1(g−1)f2(h). (41.4)

We think of both sides as a function of h; thus gf =∑f1(g−1)f2.

Now if U is left-invariant, i.e. U(gf) = gU(f), we get

U(gf)h =∑

f1(g−1)U(f2)(h) = U(f)(g−1h) (41.5)

Let’s take h = e. Then∑f1(g−1)U(f2)(e) = U(f)(g−1), and so

U(f) =∑

f1 U(f2)(e). (41.6)

Hence the functional λ : f 7→ U(f)(e) in O(G)∗ determines U , and U(f) ∈ O(G).

Now we let U be a left-invariant vector field, i.e. a derivation on O(G). Then λ is a point derivation:λ(fg) = f(e)λ(g) + λ(f) g(e) for f, g ∈ O(G).

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Let U, V be two derivations. Using (41.6), we see that UV f =∑U(f1)V (f2)(e), and so λUV : f 7→

(UV f)(e) is actually∑U(f1)(e)V (f2)(e) =

∑λU (f1)λV (f2).

So we get a product O(G)∗ ⊗ O(G)∗ → O(G)∗ by λ, µ 7→ λµ, defined by the pairing 〈λµ, f〉 def=〈λ ⊗ µ,∆f〉. This is an associative prodoct, and ε : O(G) → C determines a unit element inO(G)∗.

So we get an algebra homomorphism U(g) ↪→ O(G)∗, which takes g ↪→ point derivations at e. Sog lives in O(G)∗. But G does as well. If g ∈ G, then evg : O(G) → C is a linear map, and so thegroup algebra C[G] is also a subalgebra of O(G)∗. In fact, there’s a lot more, and O(G)∗ is a hugealgebra.


**I was 15 minutes late.** Announcement: HW is due not next Wednesday, but the Wednesdayafter that (Dec. 17).

We recall is an affine algebraic group G over C is a finitely-generated Hopf algebra O(G). We recallthat then O(G)∗ is naturally an algebra containing C[G] and U(g). If Gy V is a finite-dimensionalalgebraic G-module, then the action gives a coaction V ∗ → O(G)⊗ V ∗.

V ∗ O(G)⊗ V ∗

O(G)⊗ V ∗ O(G)⊗O(G)⊗ V ∗

coact

coact

id ⊗ coact

comult ⊗ id

There are left coactions and right coactions, and if G acts on the left on V , then O(G) coacts onthe left on V ∗, but G acts on the right on V ∗. In any case, we have an action O(G)∗ on V , and itspecializes to the actions Gy V and U(g) y V .

So, we raise the following question: given a Lie algebra representation, does it integrate to analgebraic action? We will construct O(G) from U(g), by finding each within the space of linearoperators on the other.

So, let g be a finite-dimensional Lie algebra over C. We have no hope of picking out G since g doesnot determine G: we need more data. This data will come from the representation theory of G. So,let M me a **full subcategory of the** category of finite-dimensional g-modules, containing 0and 1 and closed under ⊕, ⊗, and ()∗.

Now, let A ⊆ U(g)∗ be the set of maps of the form U(g)→ EndVφ→ C for V ∈M, where the map

φ : EndV → C is in (EndV )∗ ∈M. We think of A as the set of “” of V as V ranges.

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Dualizing, we have a map (EndV )∗ → U(g)∗. Of course, U(g) is naturally a cocommutative (but notusually commutative) Hopf algebra: if x ∈ g, then ∆x = 1 ⊗ x + x ⊗ 1 (x is primitive), Sx = −x,and εx = 0. Anyway, then U(g)∗ is naturally a commutative algebra, with the multiplication ·defined by by 〈λ · µ, u〉 def= 〈λ⊗ µ,∆u〉.

Why did we pick this Hopf algebra structure? Because it agrees with the action of g on tensorproducts. Indeed, if f, g ∈ U(g)∗ are matrix entries of V,W , then fg is a matrix entry of V ⊗W .So A is closed under multiplication, and it turns out (it’s not completely obvious **just uses thatM is closed under ⊕?**) that A is also closed under addition. Scalar multiplication is obvious,so A is a subalgebra of U(g)∗. Also, S provides a map A → A since M is closed under dualizing.which a priori is an opposite algebra homomorphism, but ∆ is cocommutative. Also, ε ∈ A is theunit element.

Anyway, we’d like to make A into a full Hopf algebra, where the coproduct should come from theproduct. Well, if U⊗U → U , then dualizing gives U∗ → (U⊗U)∗, but in general U∗⊗U∗ ↪→ (U⊗U)∗

but is not all of it. If U is finite-dimensional, then we do get a coproduct. Well, in fact any givenelement of A is a matrix coefficient for some finite-dimensional V , and so we can use the coproductconstruction with the finite-dimensional EndV . Thus, A has a coproduct ∆ dual to the product inU(g). SInce A ⊆ U(g)∗, we have a pairing, i.e. a map U(g)→ A∗ which is an algebra homomorphism,where A∗ has an algebra structure from ∆ on A. In general, this map is not injective, e.g. ifM is just(direct sums of) the trivial representation. But U(g) ↪→ A∗ if M contains a faithful representationof g.

Let’s make another assumption onM: that there exists a generator V0 ∈M such that every V ∈Mis a subquotient of some

⊕i V⊗mi

0 for some integers mi. Then (EndV0)∗ generates A as an algebra.Question from the audience: What’s a subquotient? Answer: A submodule of a quotient. Inthe one example, everything will occur as a summand, but we don’t need this. **Is it clear thatM is closed under subquotients?** Anyway, then A is a finitely-generated commutative Hopfalgebra, and hence O(G) for some algebraic group G.

It’s not clear that G has the right Lie algebra, and in fact it won’t always. But G does act oneach V ∈ M. Indeed, let v1, . . . , vn be a basis of V and ξ1, . . . , ξn the dual basis of V ∗ **I haveraised some indices**. Then we have σ : V → V ⊗ O(G) given by σ(v) =

∑vi ⊗ λi, where

λi(U) = 〈ξi, Uv〉 for U ∈ U(g). Then each λi is a matrix coefficient: λi ∈ A. Thus we have whatwe want, and Uv =

∑viλi(U), and so σ is a coaction of O(G) on V , hence an action of G and also

of U(Lie(G)) (remember that Lie(G) is not necessarily g).

Ok, so we have σ : G y V and U(Lie(G)) y V . How does Lie(G) y V ? By σ contracted withpoint derivations. But g y V also. Well, we had U(g) → A∗ = O(G)∗, and x ∈ g maps to apoint-derivation under this, since x is primitive in U(g). I.e. g→ O(G)∗ factors:

g //

��

O(G)∗

��Lie(G) // gl(V )

(42.1)

121

In any case, if M contains a faithful g-module, then g → Lie(G) is injective (it never depends onV , coming just from U(g) → A∗. But it generally is not onto. For example, let’s take g = C andM generated by Vα and Vβ, where the generator x ∈ g acts by α on Vα = C and by β on Vβ = C,and we’ve picked α, β ∈ C with α 6∈ Qβ. Then M is in fact generated by Vα ⊕ Vβ, so x acts byÇα

β

å. Well, tensor-product inM in fact is commutative, and Lie(G) will contain all diagonal

matrices since α/β 6∈ Q. But g will embed one-dimensionally. The picture is the irrational line inthe torus; of course, this is a complexification of that.

Anyway, the relationship will be that G is the Zariski closure — the smallest subvariety containing— of exp(g) ⊆ GL(V ). So the problem will be how to identify when this is Zariski-closed.


**I missed the first half. Bullets are cribbed from Alex Fink.**

• Reminder from last time. Let g finite-dimensional Lie algebra over C and M a categoryof modules over it, and M has a generator V0. We constructed a linear algebraic groupG ⊆ GL(V0) so that O(GL(V0))� O(G). We have

g //

!!BBBBBBBBB Lie(G)

zzuuuuuuuuu

gl(V )

(43.1)

and g ↪→ Lie(G) if M contains a faithful module.

• g is algebraically integrable with respect to M iff g→ Lie(G) is an isomorphism. You shouldthink ofM as the category of f.d. reps of G, and it actually is up to some closure of something.

• We have a pairing U(g) ⊗ O(G) → C, with some geometry: if H ≤ G is a Lie subgroupwith Lie(H) = g, and f ∈ O(G) and U ∈ U(G), then 〈U, f〉 = U(f |H)(e). This constructiondepends only on a neighborhood of e, so depends only on f |expW for W a neighborhood of0 ∈ g. If f ∈ O(GL(V0)), look at its image f ∈ O(G); then through some magic, we concludethat expW is Zariski-dense in G.

• Theorem 43.1: If g =⊕

gi as vector-spaces, and if each gi is algebraically integrable, thenso is g.


Integrate each gi to Gi. Then look at the map m : G1 × · · · × Gr → G, which justmultiplies in the given order. This is not a group map, just a map of algebraic varieties.Then this factors through H, because Gi → G factors through H, and H is closed (quagroup). Then the differential at the identity is just

⊕gi → g. So it has image containing

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a neighborhood of 0, and so m has image containing a neighborhood of the identity inG. Since this neighborhood is dense, m is a dominant morphism (i.e. a morphism ofalgebraic varieties with Zariski-dense image).

Thus dimG ≤ dimG1 × · · · ×Gr = dim g, so g = Lie(G). �

• The torus example. If g is abelian, then we want to integrate it to a torus. We take a latticeX ∈ g∗ of full rank (meaning X ⊗Z C = g∗); ξ1, . . . , ξn a Z-basis. We take M = {⊕Cλ : λ ∈X}, where g y Cλ by g 7→ λ(g)×. Then V0 =

⊕Cξi is a faithful generator.

Thus we have G ⊆ GL(V0) the Zariski-closure of exp g, and exp(~z) is the diagonal matrixwhose i, i entry is eξi(~z). So G is a torus T ∼= (C×)n, with O(T ) = C[t±1

1 , . . . , t±1n ], where ti of

a diagonal matrix is the (i, i)th entry.

What are the modules? If λ =∑aiξi for ai ∈ Z, then T y Cλ by (t1, . . . , tn) 7→ tα1

1 . . . tαnn .

In any case, then O(T ) = C[X], and

HomAlgGp/C(T,C×) = HomLieGp/C(T,C×) = HomAlgVar/C(T,C×)/scalars,

and T = g/2πiX∗.

We come to our main example. Let g be semisimple over C. We know its representation theory,and we know how to classify the semisimple gs. We want to know all its Lie groups.

We make some choices: h = g0 its Cartan, g =⊕α gα, and root and weight lattices Q and P . We

pick a lattice X between these Q ⊆ X ⊆ P . LetM be the finite-dimensional reps of g such that thehighest weight is in X; then all weights will be in X since X ⊇ Q. SoM = {⊕Vλ s.t. λ ∈ P+∩X}.This is a perfectly good category with ⊗ and duals.

All of this is a little abstract: to actually get your hands on the algebraic group from this descriptionis a little involved.

So, what happens to h? h acts diagonally, so it’s algebraically integrable, integrating to a complex“torus” T ∼= (C×)n with character lattice X. What about the n+ and n− parts? They’re nilpotent,and we claim that this means they’re automatically algebraically integrable. Because if g y Vnilpotently, then g ↪→ strictly upper triangular matrices. And we can exponentiate a strict-upper-triangular to get an upper-uni-triangular matrix, and exp(nilpotent) is a polynomial map. By thesame token, log(1 +X) is polynomial for X nilpotent. So we have polynomials making the strict-upper-triangular matrices isomorphic as an algebraic variety to the upper-uni-triangular matrices.But g is a vector subspace of the strict upper triangulars, so exp : g→ G makes G into an algebraicvariety.

So h, n+, and n− are algebraically integrable, and g = n− ⊕ h ⊕ n+, integrates to some G withLie(G) = g, and G ⊆ GL(Vλ) algebraically.

Question from the audience: What’s the generator of M? Answer: Ah, I didn’t show youthat M has a generator. But it does: the intersection of X with the dominant cone is a cone inthe lattice. And a cone in a lattice has a finite set of generators which generate it as a semigroup.

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And Vλ⊗Vµ = Vλ+µ⊕ other bits. So pick some generators λi, and then⊕Vλi generates. So if you

want G ⊆ GL(⊕Vλi).

So we get these different groups. We want to show that this is all of them. Well, we know thatthere is a simply-connected one, and that any other is this mod something in the center. So weneed to see that if X = P we get a simply-connected group, and if X = Q then we’ve killed thewhole center. Then the center of the simply-connected one is P/Q, and so any subgroup of thecenter is a lattice between P and Q.

So, another fact about algebraic geometry. We have g = Lie(G), and U± = exp(n±), T = exp(h),and U− × T × U+ → G with Zariski-dense image.

Oh, by the way, G is connected. It’s an algebraic variety, and these are either irreducible orreducible, which means that O(G) is either an integral domain or it isn’t. But in fact G is connectedi.e. irreducible, because it’s the Zariski closure of expW for W a neighborhood of g.

Anyway, the image U− × T × U+ → G is Zariski dense, and contains a Zariski-open set. This isanother algebraic geometry fact: the image of an algebraic map contains a set Zariski-open in itsZariski-closure.

Ok, so the complement of the image lives inside some closed subvariety of G, which will havecomplex co-dimension at least 1, and hence real co-dimensional at least 2. Because locally we’re atthe vanishing set of some polynomial in Cn. So in any one-complex-dimensional slice transverse tothe locus, the locus is just some points. Anyway, so that means that if we have any path in G, wecan move it off this closed subvariety. This really is using that real codimension is at least 2.

Anyway, so let’s take a path in G from e to e. Then up to homotopy we can get it off the complementof the image. So we have a map π1(U−TU+)� π1(G). On the other hand, U−TU+

∼← U−×T ×U+

by LU-factorization. So π1(U−TU+) = π1(U−× T ×U+) = π1(T ), since U± is affine. And π1(T ) =X∗ is the co-lattice to X, i.e. the points in g on which all of X takes integral values.

If α∨i = hI ∈ X∗, then we take sl(2)i ⊆ g and exponentiate to SL(2,C) → G. So the circlesexp(Rhi) that generate π1(T ) go to circles in SL(2,C) which is simply-connected. So the mapπ1(T )� π1(G) kills these circles, and G is simply-connected.


We begin with course evaluations.

44.1 Finishing up

Remember where we were. We have g a Lie algebra, M a good category of representations of g,and from this construct a group G. Not we let g be semi-simple over C, M the finite-dimensionalreps with weights in X, for X a lattice Q ⊆ X ⊆ P .

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We saw last time that if X = P , then G = Gsc is simply-connected. Let’s remember how we got Gsc.We split g = n−⊕h⊕n+, and n± exponentiates polynomially, h T = (C×)n, and U− ·T ·U+ → Ghomeomorphically and indeed is algebraically isomorphic to its image. But the image contains aZariski open subset, and indeed is Zariski open, although that takes more argument, and so anypath in G can be moved into U− · T · U+ → G, where it’s contractable.

Now, let X = Q, and then Vθ = (g, ad) has weights in R ∪ {0} ⊆ Q, and generates M. Thistakes a few more basic facts about what submodules appear in Vθ ⊗ Vnθ for n ≫ 0. Then theconstruction gives Gad = Int(g) = exp(ad g). Remember how Ad : Gad y g; Ad(g) y g by thedifferential of Ad(g) y G at the identity. So the center of G acts trivially, but Gad y g is faithful,so Z(Gad) = 1.

We had the above decomposition Gsc ⊇ U−TscU+, and similarly a decomposition Gad ⊇ U−TadU+.And then Tsc (which of course isn’t simply-connected, but whatever) is h/2πiQ∨, and Tad =h/2πiP∨, and Tsc → Tad with kernel P∨/Q∨. Using the algebraic geometry, we know that anyZariski-open in the irreducible Gsc intersects every other open, and Gsc → Gad is a finite cover.Thus Z(Gsc) = ker(Tsc → Tad) = P∨/Q∨.

So now we know the whole story. To each root system g, possibly disconnected, and each Q ⊆X ⊆ P , we get a group G with the correct Lie group. And then M is exactly the category ofG-modules, and so we know all the groups and their representation theory. We remark that theindex [P : Q] = detA is the determinant of the Cartan matrix.

Example: This is all fairly abstract. Let’s work an example. For example, G = SO(2n + 1,C)and g = so(2n + 1,C). It’s most convenient to take the symmetric form that’s preserved by

G to be anti-diagonal:

1

. ..

1

. Then the diagonal matrices are a maximal torus, and

g = {X = −XR}, where XR is the weird anti-transpose. So

h =

z1

. . .zn

0−zn

. . .−z1

(44.1)

Then R+ = zi ± zj for i < j and 2zi. And you can verify that α∨i = εi − εi+1, αi = zi − zi+1,and α∨n = εn, αn = 2zn. This is Langlangs-dual to Bn, which is a very subtle relationship.

In any case, P∨ = Zn, and Q∨ is the sublattice {(a1, . . . , an) s.t.∑ai is even}, and Q = Zn¡

and P = Zn ∪Ä(1

2 , . . . ,12) + Zn

ä.

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So, what is the torus? It’s

T =

t1. . .

tn0

t−1n

. . .t−11

(44.2)

with exp : h→ T . The kernel is ker(h→ T ) = 2πiZn = 2πiP∨. This is the largest is can be,so Z(SO(2n+ 1)) = 1 and this is the adjoint representation.

But then there must be another group, since Q has index 2. The simply-connected coveris a two-to-one cover of SO(2n + 1), called Spin(2n + 1,C). By the way, the same thinghappens in the even case, although the index is four: it turns out that SO(2n) already hasa two-element center. It’s smallest finite-dimensional representation is V( 1

2,..., 1

2). The Weyl

group is W = Sn n {±1}n. So dimV( 12,..., 1

2) = 2n. This is the Clifford module, which is a

lovely geometric construction. You can construct Clifford algebras from the bilinear form,but actually Spin acts on the Clifford algebra, not just SO.

We’d now like to move to a slightly more general theory. We say that G a complex Lie group isreductive if it is linear and its finite-dimensional representations are completely reducible. Why dowe say it’s linear? Well, you could take e.g. an elliptic curve, which is a complex Lie group which isalso compact. But a holomorphic function on a compact thing is constant, so an elliptic curve hasonly trivial finite-dimensional representations. THere are more general examples, and the point isthat in these the fin-reps don’t see the group structure.

Anyway, so say G is reductive, and g = Lie(G), which is then necessarily reductive as a Lie algebra.Then g = gss⊕ z where z is the center and gss is semisimple, and h = hss + z. Anyway, so we repeatthe representation-theory games. We have Q∨ ⊆ hss, and we pick a lattice h ⊇ X∗ ⊇ Q∨ such thatX ⊇ Q, i.e. 〈X∗, P 〉 ⊆ Z. And we demand that X∗ be spanning: h = C ⊗Z X

∗. Then we let Mbe the finite-dimensional representations on which h acts diagonalizably (we have to impose thisas a requirement, since the center need not do it) with weights in X. Then we get G ⊇ T = exp(h)with lattice X(T ) = X. The whole story goes through, and we won’t justify all of it. The primaryingredients we have; the only extra part is that if X is too small, then G will include a copy of(C,+), which is not completely reductive, and if X contains too many vectors, then G will containa compact variety and have no fin-reps. The conditions we have make X just the right size. Weremark that a reductive abelian group is (C×)n.

Anyhoo, so the data we have includes X,X∗, and we pick vectors and dual vectors α1, . . . , αn andα∨1 , . . . , α

∨n , and we get a matrix 〈αj , α∨i 〉, and this will be a Cartan matrix of finite type. This is

called the root datum.

Example: Let X = Zn and X∗ = Zn with the obvious pairing, but take the sl root lattice

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αi = ei− ei+1 and α∨i the same. Then we get sln with an extra dimension: g = gln = sln⊕C.Then the torus is T = (C×)n, and O(T ) = C[t±1 , . . . , t

±n ], and TSLn ↪→ T , so O(TSLn) is a

quotient of O(T ), and it’s O(TSLn) = C[t±1 , . . . , t±n ]/〈t1 . . . tn − 1〉.

Gah, we have no minutes left, but I have to say one more thing. There’s a beautiful theorem whichwe don’t have time to do:

Theorem 44.1: The root data (X,αi ∈ X,α∨i ∈ X∗) exactly classify:

• reductive Lie groups over C

• reductive algebraic groups over C

• compact real Lie groups Gθ

(We could do all this story over R, but you won’t get a compact thing. On the otherhand, inside C you can combine the Weyl reflections with complex conjugation to getan automorphism of each sl(2) whose fixed points are an so(3,R).)

• reductive algebraic groups over any algebraically closed field in any characteristic

• “algebraic groups over Z”

(We could do all this story over Z, constructing a Hopf algebra over Z, called a groupscheme, and tensoring with any field gives a real group. Tensoring with a finite groupgives a finite Chevellay group.)

• Finite groups of Lie type, which come in the same way as Gθ — use automorphisms —but over a finite group, and this is basically the source of the finite simple groups.

Theo’s answers to Problem Set 1

1. (a) Show that the orthogonal groups On(R) and On(C) have two connected components, theidentity component being the special orthogonal group SOn, and the other consisting oforthogonal matrices of determinant −1.

**I will use “1” for the identity of any group. Let K = R or C. We considerthe K-valued dot-product o Kn given by 〈x, y〉 = xT y =

∑ni=1 xiyi.

Certainly there are at least two connected components of On: if XTX = 1,then detX2 = 1, so detX = ±1; each of “detX = +1” and “detX = −1” is aclosed condition. So it suffices to show that SOn = {X ∈ On s.t. detX = 1} isconnected. We remark first-of-all that SO1 = {1} is trivially connected.

The Graham-Schmidt procedure, by which a basis is made into an orthonor-mal basis, is continuous.1 In particular, given an orthonormal basis of Kn, we

1When K = C, with the dot-product we have chosen, Graham-Schmidt can fail to turn a basis into an orthonormal

127

can perturb the first coefficient slightly and then construct an orthonormalbasis containing the perturbed first coefficient that is “near” the originalbasis (there’s a δ depending on n and ε so that any perturbation of the firstbasis vector by δ perturbs the rest by ε).

If n ≥ 2, then the unit vectors in Kn are path connected.2 Given two elementsX,Y ∈ SOn, let x1, . . . , xn and y1, . . . , yn be the columns of X and Y respectively;then each is an orthonormal basis of Kn. We pick any path z(t) from z(0) = x0

to z(1) = y0, and use the continuity of Graham-Schmidt to find a path Z(t) ∈SOn such that Z(0) = X and Z(1) has y1 as its first column. But then thecolumns {z(1)2, . . . , z(1)n} are an orthonormal basis of (y1)⊥, with the sameorientation as {y2, . . . , yn}. By induction, we can find a path in SOn−1.**

(b) Show that the center of On is {±In}.

**Let Z ∈ On be central, and let x be a unit vector (xTx = 1 ∈ K). LetX : y 7→ −y + 2(xT y)x be the reflection through x. Then XZx = ZXx = Zx,but if Xy = y, then 2y = 2(xT y)x, so y is parallel to x. Thus every vectoris an eigenvector of Z, and so Z is multiplication by a constant. The onlyconstants in On are ±1.**

(c) Show that if n is odd, then SOn has trivial center and On ∼= SOn × (Z/2Z) as a Liegroup.

**If n is odd, then the reflection X used in the previous answer is in SOn,and so the central elements are only ±1, but −1 6∈ SOn when n is odd. AsSOn has index 2 in On, it’s normal, and we have a split exact sequence

1 // SOn� � // On // Z/2Z

tvv

// 1 (PS1.1)

where t sends the generator to −1, which acts trivially on SOn.**

(d) Show that if n is even, then the center of SOn has two elements, and On is a semidirectproduct (Z/2Z) n SOn, where Z/2Z acts on SOn by a non-trivial outer automorphismof order 2.

** The split exact sequence

1 // SOn� � // On // Z/2Z

tvv

// 1 (PS1.2)

basis, because there are non-zero vectors of zero-length. But if a basis is close to being orthonormal, then Graham-Schmidt will succeed.

2For example, if∑

(xi)2 = 1 =

∑(yi)

2, then zi(t) =√

(xi)2(1− t) + (yi)2t is a path of unit-length vectors. IfK = C, then we have to pick branches of the square root; but (zi)

2(t) is an affine function of t, so has at most oneroot in t ∈ [0, 1], so there are at most n times when a choice of branches is involved, and of course any choice works.

128

now sends the generator Z/2Z to

T =

−1

1. . .

1

(PS1.3)

Then detT = −1, and of course On = (Z/2Z) n SOn where the action is byconjugation by T . We must show that this is an outer automorphism of SOn.If there were some S ∈ SOn such that for every X, TXT−1 = SXS−1, then inparticular S−1TXT−1S = X, and so S−1T commutes with all of SOn. But thenS−1T = ±1 ∈ SOn, but T 6∈ SOn.

So it suffices to show that ±1 are the only matrices Z ∈ On that commutewith all X ∈ SOn. This is false when n = 2, in which case SO2 = U(1) is thecircle group, and all elements are central. But then there are no nontrivialinner automorphisms, and O2 is a noncommutative semidirect product of twocommutative groups.

Let n > 2 is even, on the other hand, and let Z ∈ On commute with SOn.Given x, construct the reflection X as above; then Z commutes with TX:ZTX = TXZ. But then T−1ZT commutes with X for any reflection X througha line, in which case T−1ZT must equal ±1 since takes every x ∈ Kn as aneigenvalue. Thus Z = ±1, completing the proof. **

2. Problems 5–9 in Knapp Intro §6, which lead you through the construction of a smooth grouphomomorphism Φ : SU(2) → SO(3) which induces an isomorphism of Lie algebras andidentifies SO(3) with the quotient of SU(2) by its center {±I}.

**Having not yet received my copy of Knapp from Amazon, I will interpret thisexercise as “construct an isomorphism SU(2)/{±1} ∼→ SO(3).” I remember seeinga solution in the Quantum Field Theory textbook by Peskin and Schroeder, sohow hard can it be?

We begin by investigating the real Lie algebras so(3) = {A ∈M3(R) s.t. A+AT = 0}and su(2) = {a ∈ M2(C) s.t. a + a∗ = 0, tr a = 0}. We can find explicit R-bases forso(3) and su(2):X =

0 0 00 0 10 −1 0

, Y =

0 0 10 0 0−1 0 0

, Z =

0 1 0−1 0 00 0 0

⊆ so(3) (PS1.4)®

x =ñi 00 −i

ô, y =

ñ0 1−1 0

ô, z =

ñ0 ii 0

ô´⊆ su(2) (PS1.5)

These satisfy [X,Y ] = Z, [Y,Z] = X, [Z,X] = Y , and [x, y] = z, [z, x] = y, and [y, z] = x.So the Lie algebras are isomorphic.

129

We continue our explicit calculations, using the embedding of C ↪→ M2(R) : i 7→ñ0 1−1 0

ôto embed M2(C) ↪→M4(R):

etY =

1 0 00 cos t sin t0 − sin t cos t

(PS1.6)

etZ =

cos t 0 sin t0 1 0

− sin t 0 cos t

(PS1.7)

etX =

cos t sin t 0− sin t cos t 0

0 0 1

(PS1.8)

etx =ñeit 00 e−it

ô=

cos t sin t 0 0− sin t cos t 0 0

0 0 cos t − sin t0 0 sin t cos t

(PS1.9)

ety =ñ

cos t sin t− sin t cos t

ô=

cos t 0 sin t 0

0 cos t 0 sin t− sin t 0 cos t 0

0 − sin t 0 cos t

(PS1.10)

etz =ñ

cos t i sin ti sin t cos t

ô=

cos t 0 0 sin t

0 cos t − sin t 00 sin t cos t 0

− sin t 0 0 cos t

(PS1.11)

Equations (PS1.6–PS1.8) give the canonical action of SO(3) on R3: etX is therotation by t radians around the e1-axis, etc. This action defines SO(3). Equations(PS1.9–PS1.11) give the canonical action of SU(2) on R4 ∼= C2. Thus SU(2) alsoacts on

∧2(R4) ∼= R6. We compute this action, picking a particularly nice basis:

f1 = e1 ∧ e2 − e3 ∧ e4 (PS1.12)f2 = e1 ∧ e3 + e2 ∧ e4 (PS1.13)f3 = e1 ∧ e4 − e2 ∧ e3 (PS1.14)g1 = e1 ∧ e2 + e3 ∧ e4 (PS1.15)g2 = e1 ∧ e3 − e2 ∧ e4 (PS1.16)g3 = e1 ∧ e4 + e2 ∧ e3 (PS1.17)

130

In this basis, we can evaluate explicitly the action of SU(2) y R6:

etx∣∣∣R6

=

1 0 0 0 0 00 cos(2t) − sin(2t) 0 0 00 sin(2t) cos(2t) 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

(PS1.18)

ety∣∣∣R6

=

cos(2t) 0 sin(2t) 0 0 00 1 0 0 0 0

− sin(2t) 0 cos(2t) 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

(PS1.19)

etz∣∣∣R6

=

cos(2t) − sin(2t) 0 0 0 0sin(2t) cos(2t) 0 0 0 0

0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

(PS1.20)

This is all to say that the action SU(2) y R6 fixes the subspace spanned by{g1, g2, g3}, and acts as SO(3) on the subspace spanned by {f1, f2, f3}. In particular,etx acts as e−2tX , ety as e−2tY , and etz as e−2tZ .

This, then, gives us the desired homomorphism. To wit: As t varies, the circlesetx, ety, and etz generate SU(2). We can construct a homomorphism out of SU(2)by specifying where each generator goes, and the homomorphism exists if thespecification respects any relations between generators. But since SU(2) acts onR3, the map etx 7→ e−2tX , etc., from SU(2) 7→ SO(3) necessarily respects all requisiterelations: SO(3) is defined by its action on R3. This homomorphism is onto, sincee−2tX et al. generate SO(3).

The homomorphism is not one-to-one. In particular, eπx = −1 ∈ SU(2) maps toe−2πX = +1 ∈ SO(3). But {±1} ⊆ SU(2) is exactly the kernel: let a ∈ SU(2) ≤SO(4) ≤ U(4) act trivially on R6 =

∧2R4, and diagonalize a over C4: a(ei) = αiei forα ∈ C. We have implicitly tensored everything with C. Since ei ∧ ej = a(ei ∧ ej) =αiαjei ∧ ej, we must have αiαj = 1 for i 6= j. But then αj = αk = α for all j, k, andα2 = 1. Thus a acts on C2 ∼= R4 by multiplication by α = ±1. **

3. Construct an isomorphism of GL(n,C) (as a Lie group and an algebraic group) with a closedsubgroup of SL(n+ 1,C).

**We map X ∈ GL(n) toñX 00 detX−1

ô∈ SL(n+1,C), which is a polynomial in the

131

coefficients of X and in detX−1, and in particular is s/a/h. It’s plainly one-to-onewith closed image. **

4. Show that the map C∗ × SL(n,C) → GL(n,C) given by (z, g) 7→ zg is a surjective ho-momorphism of Lie and algebraic groups, find its kernel, and describe the correspondinghomomorphism of Lie algebras.

** An inverse to the map is provided by X 7→ (detX, (detX)−1X) ∈ C× × SL(n,C).The kernel of the first map is all pairs (z, g) so that det g = 1 and zg = 1. I.e. g ∈ SLis multiplication by a constant (by z−1), and 1 = det g = z−n. Thus z is an nth rootof unity: the kernel is the cyclic group of order n.

As the kernel is discrete, the Lie algebras are isomorphic. Indeed, this followssimply by dimension count, as the map is onto and the domain and range havethe same dimension 1 + (n2 − 1) = n2. **

5. Find the Lie algebra of the group U ⊆ GL(n,C) of upper-triangular matrices with 1 onthe diagonal. Show that for this group, the exponential map is a diffeomorphism of the Liealgebra onto the group.

** We recall that if V ⊆ Mn is the space of upper-triangular matrices with 0on the diagonal and x ∈ V , then xn = 0. Thus log : U → V that sends 1 + x 7→∑n−1k=1(−1)k−1xk/k and exp : V → U sending x 7→∑n−1

k=0 xk/k! are polynomials, and it’s

easy to see that they are inverses of each other. Thus V = Lie(U), and V and Uare not just diffeomorphic, but related by inverse polynomials. **

6. A real form of a complex Lie algebra g is a real Lie subalgebra gR such that that g = gR⊕ igR,or equivalently, such that the canonical map gR⊗RC→ g given by scalar multiplication is anisomorphism. A real form of a (connected) complex closed linear group G is a (connected)closed real subgroup GR such that Lie(GR) is a real form of Lie(G).

(a) Show that U(n) is a compact real form of GL(n,C) and SU(n) is a compact real formof SL(n,C).

**We have Lie algebras u = {X ∈ Mn(C) s.t. X + X∗ = 0}, gl = Mn(C), su ={X ∈ Mn(C) s.t. X + X∗ = 0, trX = 0}, and sl = {X ∈ Mn(C) s.t. trX = 0}. Byfor any Z ∈Mn, we can write Z = X+iY uniquely with X+X∗ = Y +Y ∗ = 0, byX = (Z−Z∗)/2i and Y = (Z+Z∗)/2i. If trZ = 0, then trZ∗ = 0, so trX = 0 = trY .This expresses gl = u⊕ iu and sl = su⊕ isu.

That U (SU) is a subgroup of GL (SL) is obvious. We prove that U(n), andhence any closed subgroup (e.g. SU(n)), is compact:

If we know operator norms, the compactness of U(n) is trivial — for com-pleteness, we verify the steps. Define ‖X‖ = supv∈Cnr{0} ‖Xv‖/‖v‖, where by‖v‖ we mean the Euclidean norm of v ∈ Cn ∼= R2n. It’s enough to allow v in thesup to range over the unit sphere; for any X ∈Mn(C), v 7→ ‖Xv‖/‖v‖ is contin-uous, and so hits its maximum, and ‖X‖ < ∞. If X ∈ U(n), then ‖Xv‖ = ‖v‖

132

by diagonalizing X (any matrix X can be written in upper-triangular form,but then X−1 is upper triangular and X∗ is lower-triangular, so if X ∈ U(n),then it is diagonalizable, and its eigenvalues must be on the unit-circle inC). Moreover, for any matrix X ∈ Mn(C) and (complex) scalar α, we have‖αX‖ = |α| ‖X‖, and if ‖X‖ = 0, then ‖Xv‖ = 0 for all v, so Xv = 0 so X = 0.In any case, take any topological disk in Mn(C) = R4n2

that contains 0 in itsinterior. As ‖ · ‖ is continuous on Mn and non-zero on the boundary of thisdisk, it must take its minimum; hence, by multiplying by a large enough(positive real) scalar, we can assume this disk contains all matrices of norm1, and in particular it must contain U(n). Thus U(n) is a closed subset of acompact space, and hence compact. **

(b) Show that SO(n) is a compact real form of SO(n,C).

**We have Lie algebras so(K) = {X ∈ Mn(K) s.t. X + XT = 0} for K = R orC. We have Mn(C) = Mn(R) ⊕ iMn(R) component-by-component, and X 7→XT preserves this decomposition. That SO(n,R) is compact follows from itsembedding as a closed subgroup of the compact SU(n). **

(c) Show that Sp(n) is a compact real form of Sp(n,C).

**Our Lie algebras are sp(K) = {X ∈ M2n(K) s.t. XJ + JXT = 0} where J =ñ0 1−1 0

ôin block form (1 here is the identity matrix in Mn. We can again

use the decomposition component-by-component of Mn(C) = Mn(R)⊕ iMn(R);the linear relation XJ + JXT = 0 is preserved by complex conjugation, sois preserved by the decomposition. Compactness follows from the closedembedding of Sp(n,R) into SU(m) for high enough m. **

7. Show that a closed linear group H is compact if and only if every X ∈ Lie(H) has the propertythat iX is diagonalizable (over C) and has real eigenvalues.

**Of course, Z ↪→ GL1 by n 7→ 2n× is a closed subgroup which is not compact. Theexercise must be asking about only the connected component of the identity.

In one direction, this is straightforward. If X ∈ Lie(H) has an eigenvalue λ thatis not pure-imaginary, then in some basis the first column of X is λ, 0, . . . , 0, andthe first column of etX is etλ, 0, . . . , 0. But if λ is not pure imaginary, then eRX isa closed subgroup of GLn and hence of H, which is not compact. So H cannotbe compact. Similarly, if X is not diagonalizble, then we can find a basis of X inwhich the first two columns are

X =

λ 1 ∗0 λ ∗0 0 ∗...

... ∗

(PS1.21)

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Exponentiating gives

etX =

etλ tetλ ∗0 etλ ∗0 0 ∗...

... ∗

(PS1.22)

This similarly is a closed subgroup, since tetλ scales either linearly or exponentially(depending on whether λ has real part or is pure-imaginary), and so if tetλ ≈ sesλ,then t ≈ s. Thus, if any X ∈ Lie(H) is not diagonalizable with pure-imaginaryeigenvalues, then H contains a copy of R as a closed subgroup, but any closedsubset of a compact space is compact, and R is not compact.

On the other hand, the converse direction seems nigh intractable.**


1. (a) Show that the composition of two immersions is an immersion.

**We consider the category of s/a/h manifolds with s/a/h maps as a subcat-egory of Top (with continuous maps). An immersion N ⊆ M in Man is (i) amorphism in Man (i.e. it is s/a/h), (ii) an immersion in Top (i.e. a continuousmap N ⊆M that is a monomorphism such that N has the induced topology),and (iii) such that if Z → M in Man has a pullback Z → N ⊆ M in Top, thenthis pullback is in Man.

If N ⊆ M ⊆ L are two immersions, then N → L is of course s/a/h, and animmersion in Top. We chase a diagram:

Z

s/a/h

��

cont’s

__

Z

s/a/h

��cont’s

zz

cont’s

__

Z

s/a/h

��s/a/h

zz

cont’s

__

Z

s/a/h

��s/a/h

zz

s/a/h

__⇒ ⇒ ⇒ ⇒

N ⊆ M ⊆ L N ⊆ M ⊆ L N ⊆ M ⊆ L N ⊆ M ⊆ L

The first implication is just composition; the latter two are using the univer-sal property of the immersions M ⊆ L and N ⊆M .

We remark that in both Top and Man, the functor of points Hom({pt},−) isnot faithful, and that this is the crux of the matter. The notion of “immer-sion” makes sense for any subcategory C ⊆ D, where D itself has a notion ofimmersion. To wit: N →

CM is an immersion in C if N →

DM is an immersion

in D such that any map Z →CM that factors through N in D factors in C. The

above diagram chase, with “s/a/h” and “cont’s” replaced by “arrow in C”and “arrow in D”, shows that immersions compose in any category. (More-over, if C’s immersions are relative to D’s, and D’s are relative to E’s, then a

134

similar chase makes it clear that C ⊆ E can inherit its immersions directly.)If we take “injection” to be the notion of immersion in Set, then in partic-ular immersions in Top (subspaces with induced topology) are exactly whatthey should be for the subcategory Top ⊆ Set. Since injections are exactlymonomorphisms, immersions solve the problem of the failure of faithfulnessby the functor of points. **

(b) Show that an immersed submanifold N ⊆M is always a closed submanifold of an opensubmanifold, but not necessarily an open submanifold of a closed submanifold.

**For the converse direction, we consider the spiral R → R2 given in polarcoordinates by (r(t), θ(t)) = (1 + et, t). This is certainly an immersion. Anyclosure of the image must contain the circle {r = 1} ⊆ R2, but any open setthat intersects this circle in the closure must also intersect the spiral. Aclosed submanifold of R2 containing this spiral must contain a chart near(r, θ) = (1, 0), which intersects the spiral infinitely often and so cannot be one-dimensional. Thus any closed submanifold containing the spiral is at leasttwo-dimensional, and the spiral is not open in a two-dimensional manifold.

For the forward direction, we let f : N ↪→M be an immersed submanifold, p ∈N , and p ∈ T ⊆ N a compact neighborhood: i.e. a compact set so that p is in

the interior◦T . Since T is compact, f(T ) is compact and in particular closed in

M . Since N has the induced topology, f(◦T ) = U∩f(N) for some open U ⊆

openM ,

and f(◦T ) = U ∩ f(T ) is closed in U . Repeating this for each point p ∈ N to get

Tp 3 p and Up 3 f(p), we take as our open submanifold⋃p∈N Up ⊆

openM (every

open subset of a manifold is an open submanifold). Then f(N) ⊆ ⋃U , and

has no intersection with the open subset⋃p∈N (Up r f(Tp)) ⊆

open

⋃U . On the

other hand,⋃p∈N Up ⊇ f(N) ∪ ⋃p∈N (Up r f(Tp)) =

⋃p∈N (f(N) ∪ (Up r f(Tp))) ⊇⋃

p∈N

Åf(◦T p) ∪ (Up r f(Tp))

ã=⋃p∈N Up, so f(N) is closed in

⋃p∈N Up. **

2. Prove that if f : N → M is a smooth [analytic, holomorphic] map, then (df)p is surjectiveif and only if there are open neighborhoods U of p and V of f(p), and an isomorphismψ : V ×W → U , such that f ◦ ψ is the projection on V .

In particular, deduce that the fibers of f meet a neighborhood of p in immersed closedsubmanifolds of that neighborhood.

3. Prove the implicit function theorem: a map (of sets) f : M → N between manifolds is smooth[analytic, holomorphic] if and only if its graph is an immersed closed submanifold of M ×N .

**Let f : M → N be s/a/h. Then (id, f) : M →M×N is s/a/h and one-to-one. Theprojection map πM : M ×N →M is s/a/h, and so if g : Z →M ×N hits the image(id, f)(M) is continuous, then πM ◦ g : Z → M is continuous, so the (id, f)(M) has

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the induced topology, and if g is s/a/h, then so is πM ◦ g. But (id, f) ◦ πM ◦ g = g,so (id, f) is an immersion.

Conversely, if (id, f) : M →M ×N is an immersion, then in particular it is s/a/h,and πN : M ×N → N is s/a/h, so f = πN ◦ (id, f) is s/a/h.**

4. Prove that the curve y2 = x3 in R2 is not an immersed submanifold. [This is a strongerstatement than the observation we made in class that the smooth bijection t 7→ (t2, t3) of Ronto this curve is not an immersion.]

**For want of a contradiction, let R→ R2 mapping s 7→ (x(s), y(s)) be an immersiononto {x3 = y2}. In particular, x(s) and y(s) are both smooth, and we will regularlyuse polynomial approximation. By translating s, we can assume that 0 7→ (0, 0);so x(s) = x′(0)s + x′′(0)s2/2 + x′′′(0)s3/6 + O(s4) and y(s) = y′(0)s + y′′(0)s2/2 + O(s3).The functions x(s) and y(s) satisfy (x(s))3 = (y(s))2. But (x(s))3 = (x′(0))3s3 + O(s4)whereas (y(s))2 = (y′(0))2s2 + O(x3), so y′(0) = 0. But then (y(s))2 = (y′′(0))4s4/4 +O(s5), so x′(0) = 0. But then (x(s))3 = (x′′(0))6s6/8, so y′′(0) = 0. In particular, wesee that x(s) = x′′(0)s2/2 + O(s3) and y(s) = y′′′(0)s3/6 + O(s4), where (x′′(0)/2)3 =(y′′′(0)/6)2.

So x(s) = O(s2) and y(s) = O(s3). If (x(s), y(s)) is an immersion, then the smoothfunctions of s ∈ R must be exactly the restrictions of smooth functions of (x, y) ∈R2. But the function s 7→ s : R→ R cannot be such a restriction. If it were, we’dhave S(x, y) : R2 → R smooth with S(x(s), y(s)) = s, so in particular S(x, y) = S(0, 0)+Sx(0, 0)x+ Sy(0, 0) y +O(x, y)2, where Sz = ∂S/∂z. Composing with x(s) = O(s2) andy(s) = O(s3) gives s = S(0, 0)+Sx(0, 0)O(s2)+Sy(0, 0)O(s3)+O(s4), which is impossible.**

5. Let M be a complex holomorphic manifold, p a point of M , X a holomorphic vector field.Show that X has a complex integral curve γ defined on an open neighborhood U of 0 inC, and unique on U if U is connected, which satisfies the usual defining equation but in acomplex instead of a real variable t.

Show that the restriction of γ to U ∩ R is a real integral curve of X, when M is regarded asa real analytic manifold. [This exercise is meant to clarify a point left vague in the lecture.]

**By working in a holomorphic chart on a neighborhood of p ∈M , we can assumethat X is a holomorphic vector field on a neighborhood of ~0 ∈ Cm. We let X(~w) bethe vector ~X(~w) ∈ Cm ∼= T~wCm, so each Xi : Cm → C is holomorphic, and restrictour neighborhood to be connected and simply connected in Cm. We are lookingfor a holomorphic map ~W (z) : C ⊇

openU → Cm so that dW i(z)/dz = Xi( ~W (z)) and

W i(0) = 0.

Writing everything in power series, we have a neighborhood of ~0 ∈ Cm so that

Xi(~w) =∞∑n=0

1n!∂nj1,...,jnX

i|0wj1 . . . wjn (PS2.1)

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where we adopt the physicists’ convention of summing over repeated indices, andof writing ∂nj1,...,jn for the operator ∂n/(∂wj1 . . . ∂wjn). Thinking of the terms ∂lW j |0as unknown constants to solve for, we are looking for a (actually m different)power series

W j(z) =∞∑l=1

1l!∂lW j |0zl (PS2.2)

defined in a (connected and simply-connected, indeed necessarily circular) neigh-borhood of the origin, such that

∞∑n=0

1n!∂n+1W i|0zn =

∞∑n=0

1n!∂nj1,...,jnX

i|0n∏k=1

( ∞∑l=1

1l!∂lW jk |0zl

)(PS2.3)

Since the sum inside the product starts in degree 1, we see immediately that thisdefines each ∂nW j |0 inductively: by equating degrees in z, we have ∂W i|0 = Xi|0,∂2W i|0 = Xj |0∂jXi|0, etc. Thus the curve ~W (z) is unique if it exists, and we cancompute it term-by-term in power series. We have only to show constants ∂nW i|0defined by the equation in formal power series (PS2.3) make the sum in (PS2.2)converge in a neighborhood of the origin.

But the functions Xi are analytic as real functions of the real variables Cm ∼=R2m, so there is a neighborhood of 0 ∈ R and a function V : R → Cm so thatdV i(x)/dx = Xi(~V (x)). The Taylor coefficients dnxV

j |0 of V j satisfy exactly theequations defining ∂nW j |0 in (PS2.3), and if ~V (x) is analytic, then the m sums∑∞l=1 d

lV j |0xl/l! all converge absolutely in a neighborhood of 0 ∈ R. Thus the sums(PS2.2) converge absolutely in a circle around 0 ∈ C, provided real-analytic vectorfields admit real-analytic integral curves. That the restriction of W to the realline is a real curve is immediate: V (x) = W (x+ 0i) by construction.

We now proceed to show that an analytic vector field has an analytic integral.Using the notation as above, let X : Rm → Rm be an analytic vector field. Then thepower series

∑∂nX|0wn/n! converges to Xi in a sphere around the origin, and we

will suppress vector indices, writing ∂n for the nth derivative operator. (Thus,from here on out everything is (symmetric) tensor-valued, and multiplicationsinclude the natural contractions.) By the usual remarks about Taylor series anda few applications of the AM-GM inequality, this convergence is faster in a smallenough sphere than a geometric series: there is a real number a > 0 so that‖∂nX|0‖ < n!an. Rescaling X(w) Y (w) = 1

aX(w/a), we see that ‖∂nY |0‖ < n!/a,and if W (z) is a solution to W ′(z) = X(W (z)), then U(z) = 1

aW (z) is a solutionto U ′(z) = Y (U(z)). It clearly suffices to show that if Y (w) has the propertythat ‖∂nY |0‖ < n!/a for some a, then the initial value problem U ′(z) = Y (U(z)),U(0) = 0 has an analytic solution. Our proposed solution, of course, defines U(z)inductively in power series:

∂n+1U∣∣∣0

= ∂n[Y ◦ U ]|0 =∑π∈Pn

∂|π|Y∣∣∣0

∏β∈π

∂|β|U∣∣∣0

(PS2.4)

137

The second equality is by Faa di Bruno’s formula: Pn is the set of partitions ofthe set {1, . . . , n}, π is a partition thought of as a set of blocks β, and we use |σ|for the size of a set σ. We wish to show that

∑∞n=0 ∂

nU |0 zn/n! converges in a smallenough interval faster than a geometric series; i.e. that there is a number c > 0so that for every k: ∥∥∥∂kU |0∥∥∥ < k! ck (PS2.5)

Indeed, we show by induction that∥∥∥∂kU |0∥∥∥ < 1 · 3 · · · · · (2k − 3) a−k (PS2.6)

which is obviously less than k! (2/a)k. Indeed, assuming (PS2.6) for k ≤ n, andrecalling (PS2.4) and that ‖∂nY |0‖ < n!/a, then we have

∥∥∥∂n+1U |0∥∥∥ < ∑

π∈Pn

|π|!a

∏β∈π

(1 · 3 · . . . (2|β| − 3)) a−|β| =1a

1 · 3 · . . . (2n− 1) a−n (PS2.7)

where to evaluate the right-hand-side we used that ∂n[1−√

1− 2t]∣∣∣t=0

= 1·3·(2n−3)

and ∂n[1/(1− u)]|u=0 = n!, and expanded ddt [1−

√1− 2t] = 1/(1− (1−

√1− 2t)) using

Faa di Bruno’s formula. This completes the induction step and hence the proof.**

6. Let SL(2,C) act on the Riemann sphere P1(C) by fractional linear transformationsña bc d

ôz =

(az+ b)/(cz+d). Determine explicitly the vector fields f(z)∂z corresponding to the infinites-imal action of the basis elements

E =ñ

0 10 0

ô, H =

ñ1 00 −1

ô, F =

ñ0 01 0

ôof sl(2,C), and check that you have constructed a Lie algebra homomorphism by computingthe commutators of these vector fields.

**We recall that the vector field, thought of as a differential, corresponding toX ∈ Lie(G) is given by

X[f ](z) =d

dt

∣∣∣∣t=0

f(exp(−tX)z)− f(z)t

=f(exp(−εX)z)− f(z)

ε(PS2.8)

where we use the convention that ε2 = 0 (i.e. we replace every function by itsTaylor polynomial with remainder: g(z+ε) = g(z)+εg′(z)+o(ε), and all our equationsactually have remainders that are o(ε)). Thus we compute actions by the above

138

generators:

exp(−εE) z =ñ

1 −ε0 1

ôz =

1z +−ε0z + 1

= z − ε (PS2.9)

exp(−εF ) z =ñ

1 0−ε 1

ôz =

1z + 0−εz + 1

=1

1z − ε

= z + εz2 (PS2.10)

exp(−εH) z =ñe−ε 00 eε

ôz =

e−εz + 00z + eε

= e−2εz = z − 2εz (PS2.11)

If f is differentiable and δ2 = 0, then (f(z + δ)− f(z)) = δf ′(z). So

E[f ](z) = f ′(z) i.e. E 7→ −∂z (PS2.12)

F [f ](z) = −z2f ′(z) i.e. F 7→ z2∂z (PS2.13)H[f ](z) = 2zf ′(z) i.e. H 7→ −2z∂z (PS2.14)

The structure of sl(2) is given by

[E,F ] = H, [E,H] = −2E, [F,H] = 2F (PS2.15)

On the other hand, the bracket of first-order differential operators is [f(z)∂z, g(z)∂z] =(f(z)g′(z)− g(z)f ′(z)) ∂z. (When z is vector-valued, ∂z is really a co-vector, f and gare vectors, and f ′ and g′ are matrices, and I have suppressed the contractions.)Sure enough: î

−∂z, z2∂zó

= −2z∂z, (PS2.16)

[−∂z,−2z∂z] = 2∂z, (PS2.17)îz2∂z,−2z∂z

ó=Äz2(−2)− (−2z)(2z)

ä∂z = 2z2∂z (PS2.18)

**

7. (a) Describe the map gl(n,R) = Lie(GL(n,R)) = Mn(R)→ Vect(Rn) given by the infinites-imal action of GLn(R).

**We use the notation in the previous exercise, where ε2 = 0, and adoptEinstein’s index convention. We let ∂i = ∂/∂zi, and if f(zi) is a scalar func-tion, then f ′i = ∂if is a covector. Taylor’s formula (defining the derivativecovariantly) reads:

f(zi + δi) = f(zi) + δif ′i(z) + o(δ) (PS2.19)

Any matrix Xji ∈ gl(n) gives rise to a vector field via the canonical action of

139

GL(n):

Xz[f ] = X[f ](z) =f(exp(−εX)z)− f(z)

ε(PS2.20)

=f((1− εX) z)− f(z)

ε(PS2.21)

=fÄÄδji − εX

ji

äziä− f

(zj)

ε(PS2.22)

=f(z)− εXj

i zif ′j(z) + o(εX)− f(z)

ε(PS2.23)

= −Xji zi∂j [f ] (PS2.24)

**

(b) Show that so(n,R) is equal to the subalgebra of gl(n,R) consisting of elements whoseinfinitesimal action is a vector field tangential to the unit sphere in Rn.

**Since the vector field Xz = −Xji zi∂j is linear in z, it is tangent to the unit

sphere if and only if it is tangent to every sphere.

We let gij be the standard metric on Rn, with inverse gij (so gijgjk = δki ).

Then the sphere of radius ρ is given by the ρ2-level set of r(z) = gijzizj. A

vector field Az = Aj(z) ∂j is tangent to the spheres if and only if A[r] = 0. Butr′j(z) = 2gjkzk, so a vector field A annihilates r(z) exactly when Aj(z) 2gjkzk = 0.

Thus, X ∈ M(n) gives rise to a vector field −Xji zi∂j, which annihilates r(z)

only when Xji zigjkz

k = 0 for every z ∈ Rn. In particular, for any vectors ai

and bi, we have

0 = Xji gjk(a

i + bi)(ak + bk) (PS2.25)

= Xji gjka

iak +Xji gjkb

ibk +Xji gjka

ibk +Xji gjkb

iak (PS2.26)

= 0 + 0Xji gjka

ibk +Xji gjka

jbk (PS2.27)

=ÄXji gjk +Xi

jgikäaibk (PS2.28)

In terms of the basis, this is exactly the statement that X + XT = 0. Con-versely, if Xj

i gjk +Xijgik = 0, then

0 =ÄXji gjk +Xi

jgikäzizk = Xj

i gjkzizk +Xi

jgjkzkzi = 2Xj

i gjkzizk (PS2.29)

We remark that so can be defined with respect to any metric, in which casewe define the transpose by

ÄXTäji

def= gmjXnmgni. That X ∈ so is equivalent

to saying that Xji gjk + Xj

kgik = 0, and the argument goes through exactly asabove. Moreover, the statement that so is the subalgebra of gl tangent tothe unit sphere is true over any field of characteristic 6= 2; indeed, over anycommutative ring in which 2 is not a zero divisor. **

140

8. (a) Let X be an analytic vector field on M all of whose integral curves are unbounded (i.e.,they are defined on all of R). Show that there exists an analytic action of R = (R,+)on M such that X is the infinitesimal action of the generator ∂t of Lie(R).

**Our action takes (t, p) ∈ R ×M to∫pX(t), where

∫pX is the integral curve

of X through p. Since the integral curves are unbounded, this action is well-defined, and by definition takes the infinitesimal generator ∂t to the derivativein the direction X. We showed in problem 5 that if X is an analytic vectorfield then each integral curve

∫pX(t) is analytic in t ∈ R for t in a neighborhood

of 0. In particular, since∫pX(s+ t) =

∫∫pX(s)X(t), then

∫pX(t) is analytic in t

everywhere.

We wish to show analyticity in p. By working in a local chart, we assume thatM = Rm and ~p is in a neighborhood of 0. We let ~X : Rm → Rm be analytic, andthe integral

∫~p~X(t) is given by a vector of function ~γ : R1+m → Rm, satisfying

∂

∂tγi(t, ~p) = Xi(~γ(t, ~p)) (PS2.30)

γi(0, ~p) = pi (PS2.31)

Then in particular we have the derivatives of (PS2.31):

∂

∂pjγi(0, 0) = δij and

Å∂

∂~p

ãlγi(0, 0) = 0 for l ≥ 2. (PS2.32)

We suppress the indices on X, γ, p and expand (PS2.30) as in problem 5 viathe Faa di Bruno formula, taking lth derivatives in p. We find:Å∂

∂t

ãk Å ∂∂p

ãlγ

∣∣∣∣∣(0,0)

=∑

π∈Pk−1

∑~a

Çl

~a

å Å∂

∂p

ãa0+|π|X

∣∣∣∣∣0

∏β∈π

Å∂

∂t

ã|β| Å ∂∂p

ãaβγ

∣∣∣∣∣(0,0)

(PS2.33)

Here Pk−1 is the set of partitions π of {1, . . . , k−1}, β ranges over the blocks ofπ, N = {0, 1, . . . }, ~a = (a0, . . . , aβ, . . . , a|π|) ∈ N1+|π| is required to satisfy a0 + · · ·+a|π| = l, and

( l~a

)is the multinomial coefficient l!/(a0! . . . a|π|!). This defines each

derivativeÄ∂∂t

äk Ä ∂∂p

älγ

∣∣∣∣(0,0)

inductively in terms ofÄ∂∂t

äi Ä ∂∂p

äjγ

∣∣∣∣(0,0)

for i < k

and j ≤ l, the k = 0 case being given by (PS2.32).

If X is analytic, then there is some overall constant c such that∥∥∥Ä ∂∂pänX∣∣∣0∥∥∥ <

n!cn (choose your favorite norm). We let B : R2 → R be the following analyticfunction of two variables:

B(t, p) =1c

(1−»

1− 2pc+ p2c2 + t)

141

B solves the initial value problem ∂∂tB(t, p) = 1/(1− cB), B(0, p) = p. We let bi,j

be the (i, j)th derivative of B at 0:

bi,j =Å∂

∂t

ãi Å ∂∂p

ãjB

∣∣∣∣∣0,0

It follows immediately by induction (and the triangle inequality) that∥∥∥∥∥∥Å∂

∂t

ãk Å ∂∂p

ãlγ

∣∣∣∣∣(0,0)

∥∥∥∥∥∥ < bk,l (PS2.34)

as the bs are positive and satisfy

bk,l =∑

π∈Pk−1

∑~a

Çl

~a

åca0+|π|(a0 + |π|)!

∏β∈π

b|β|,aβ (PS2.35)

Thus, we have bounded the derivatives of γ by terms of a series with positiveradius of convergence. Hence γ is analytic in a neighborhood of 0. **

(b) More generally, prove the corresponding result for a family of n commuting vector fieldsXi and action of Rn.

**Analyticity follows by the same argument as in part (a), where now γ :Rn+m → Rm satisfies

∂

∂tjγi(~t, ~p) = Xi

j(~γ(t, ~p)) (PS2.36)

γi(~0, ~p) = pi (PS2.37)

It suffices to show that a system of n commuting vector fields, all of whoseintegral curves never blow up, integrates to an action of Rn. The general-nonsense argument is that since Rn is a simply-connected Lie group, Liehomomorphisms out of Rn are determined by Lie algebra homomorphismsof Lie(Rn), and these are given by a choice of n commuting Lie algebra el-ements; hence, since the Xis commute, the map ∂i 7→ Xi is a Lie algebrahomormorphism Lie(Rn)→ Vect(M) = Lie(Diff(M)), where ∂i = ∂/∂ti and ti arethe canonical coordinates of Rn, and we can lift this algebra map to a mapof groups Rn → Diff(M).

Without using such general nonsense (indeed, the theory developed in thisclass does not particularly cover infinite-dimensional Lie groups like Diff(M)),we can still define the action of Rn on M . Let X and Y be two commuting vec-tor fields, x(t, p) the solution to the initial value problem x(t, p) = X(x(t, p)),x(0, p) = p, and y(t, p) the solution to y(t, p) = Y (y(t, p)), y(0, p) = p, wherez(t, p) = ∂

∂tz(t, p). Since X and Y commute, we have an exact infinitesimal

142

identity x(ds, y(dt, p)) = y(dt, x(ds, p)). This can be integrated to a finite iden-tity x(s, y(t, p)) = y(t, x(s, p)). Hence we can choose whatever order we want;we define xi(t, p) to solve the initial value problem xi(0, p) = p, xi(t, p) =Xi(xi(t, p)), and then define the action γ : Rn × M → M by (t1, . . . , tn, p) 7→x1(t1, x2(t1, . . . xn(tn, p) . . . )). By the commutivity, γ(~s + ~t, p) = γ(~s, γ(~t, p)), andthis is a group action of Rn yM . **

9. (a) Show that the matrixñ−a 00 −b

ôbelongs to the identity component of GL(2,R) for all

positive real numbers a, b.

** We connectñ

1 00 1

ôtoñ−a 00 −b

ôvia the path [0, 1]→ GL(2,R) given by

t 7→

ñ

cos(2πt) sin(2πt)− sin(2πt) cos(2πt)

ô, 0 ≤ t ≤ 1/2

−ñe(2t−1) ln a 0

0 e(2t−1) ln b

ô, 1/2 ≤ t ≤ 1

**

(b) Prove that if a 6= b, the above matrix is not in the image exp(gl(2,R)) of the exponentialmap.

**The eigenvalues ofñ−a 00 −b

ôare, of course, −a and −b. If expX =ñ

−a 00 −b

ôfor X ∈ gl(2,R), then its (complex) eigenvalues must exponentiate

to −a and −b, i.e. its eigenvalues must be ln a + πi + 2πim and ln b + πi + 2πinfor integers m and n. But the eigenvalues of a real 2 × 2 matrix are eitherboth real or are complex-conjugate; ln a+ πi+ 2πim is certainly not real, andln a + πi + 2πim and ln b + πi + 2πin are only complex-conjugate if ln a = ln b,which cannot happen if a 6= b ∈ R.

We remark that the failure of the exponential map to be onto is not limited

to real closed linear groups:ñ−1 10 −1

ô6∈ exp(sl(2,C)).**


1. (a) Let S be a commutative k-algebra. Show that a linear operator X : S → S is a derivationif and only if it kills 1 and its commutator with the operator of multiplication by everyfunction is the operator of multiplication by another function.

**We recall that X : S → S is a derivation if X(fg) = X(f) g + f X(g) for eachf, g ∈ S. Then [X, f ](g) = X(fg) − f X(g), and so if X is a derivation, then

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[X, f ] = X(f). Conversely, if [X, f ] ∈ S for every f ∈ S, then [X, f ] = [X, f ]1 =X(f1) − f X(1). So if X kills 1, then [X, f ] = X(f), and X(f) g = [X, f ](g) =X(fg)− f X(g), hence X is a derivation. **

(b) Grothendieck’s inductive definition of differential operators on S goes as follows: thedifferential operators of order zero are the operators of multiplication by functions; thespace D≤n of operators of order at most n is then defined inductively for n > 0 byD≤n = {X : [X, f ] ∈ D≤n−1 for all f ∈ S}. Show that the differential operators of allorders form a filtered algebra D, and that when S is the algebra of smooth [analytic,holomorphic] functions on an open set in Rn [or Cn], D is a free left S-module with basisconsisting of all monomials in the coordinate derivations ∂/∂xi.

**Since we define D as its union of subalgebras, it suffices to show thatmultiplication respects the grading. Let X ∈ D≤n and Y ∈ D≤m be differentialoperators of degree at most n and m respectively. Recall that the commutatorin an associative algebra is a (bi)derivation. Then [XY, f ] = X[Y, f ] + [X, f ]Y ∈D≤nD≤m−1 + D≤n−1D≤m ⊆ D≤n+m−1 by induction on n + m (when n + m = 0,the result is trivial). Thus XY ∈ D≤n+m. Hence D is a filtered algebra.We observe that, since the Jacobi identity assures that bracketing is a Lie(bi)derivation, [[X,Y ], f ] = [[X, f ], Y ] + [X, [Y, f ]] ∈ [D≤n, D≤m−1] + [D≤n−1, D≤m] ⊆D≤n+m−2 by induction, hence [X,Y ] ∈ D≤n+m−1.

For the second part, we let S be the algebra of s/a/h functions on an opendomain. It’s clear that any polynomial of degree at most k in the ∂/∂xiswith coefficients in S is a differential operator of degree at most k underGrothendieck’s definition. Let X be any such polynomial in the ∂/∂xis. Then

X =∞∑n=0

1n!

îî. . .îîX,xi1

ó, xi2ó, . . .ó, xinó(1)

∂n

∂xi1 . . . ∂xin. (PS3.1)

We have used the Einstein repeated-index summation convention. The n = 0bracket of X with zero xis we let be just X. If X is of degree k in the ∂/∂xis,then (PS3.1) terminates after n = k.

Conversely, if X ∈ D≤k is an arbitrary Grothendieck differential operator,then the right-hand-side of (PS3.1) terminates after n = k and agrees with Xwhen evaluated on any polynomial. It must therefore agree on any algebraicfunction: the generalized chain rule reduces to the generalized product rulefor left-composition with polynomials. But I see no reason that it shouldnecessarily agree with X on all, even transcendental, functions without as-suming some topological condition on the differential operators. Of course,it is a (presumably hard; I don’t know how to do it) theorem of Peetr’s thatany local operator on C∞ is a differential operator. In our case, it suffices toknow that any differential operator is continuous in an appropriate topologyon S (e.g. the point-wise topology), since the polynomials are dense.

144

Indeed, I believe the claim in the problem would be false were S replacedby its field of fractions. In this case, we can understand S as a (transfinite)field extension of the polynomials. Any derivation, say, of a field extends (atmost) uniquely to any algebraic extension of that field: if f is a root of apolynomial p =

∑pkx

k ∈ T [x] (for T a subalgebra of S), and X is a derivationof T [f ] (the subalgebra of S generated by T and f), then

X(f) =∑X(pk) fk∑kpkfk−1

. (PS3.2)

(By choosing generators of the vanishing ideal of f in T [x] of minimal degree,we can assure that the denominator of (PS3.2) does not vanish identically,but it might vanish at points.) Hence extensions of derivations to alge-braic extensions are uniquely determined. But on transcendental extensionsderivations can behave arbitrarily.

In any case, there are holes in the previous paragraph, because I don’tknow enough algebra to be sure in the non-PID case (and we’re in the non-Noetherian case) that (PS3.2) is in fact consistent when X is not a sum ofmultiples of ∂/∂xis. And I don’t know enough product rules to write downsimilar formulas for the extensions of higher-order differential operators. ButI expect there’s something missing.

Naturally, Wikipedia quotes Grothendieck’s definition as equivalent to thefreshman-calculus definition.

**

2. Calculate all terms of degree ≤ 4 in the Baker-Campbell-Hausdorff formula.

**We wish to calculate β(X,Y ) = log(expX expY ), where X and Y are noncom-muting variables and exp and log are the usual formal power series. To degree≤ 4, we have

expX expY =Å

1 +X +12X2 +

16X3 +

124X4 + . . .

ãÅ1 + Y +

12Y 2 +

16Y 3 +

124Y 4 + . . .

ã(PS3.3)

= 1 +X + Y +12

ÄX2 + 2XY + Y 2

ä+

16

ÄX3 + 3X2Y + 3XY 2 + Y 3

ä+

124

ÄX4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4

ä+ . . . (PS3.4)

We let β(X,Y ) = Z1 + Z2 + Z3 + Z4 + . . . where each Zi is homogeneous in X andY of degree i. (Since β(0, 0) = 0, there is no constant term.) Then, grouping by

145

degree,

expβ(X,Y ) = 1 + Z1 +Å1

2(Z1)2 + Z2

ã+Å1

6(Z1)3 +

12

(Z1Z2 + Z2Z1) + Z3

ã+Å 1

24(Z1)4 +

16

Ä(Z1)2Z2 + Z1Z2Z1 + Z2(Z1)2

ä+

12

ÄZ1Z3 + (Z2)2 + Z3Z1

ä+ Z4

ã(PS3.5)

and we compare term by term.

Z1 = X + Y (PS3.6)

Z2 =12

ÄX2 + 2XY + Y 2

ä− 1

2

ÄX2 +XY + Y X + Y 2

ä(PS3.7)

=12

(XY − Y X) (PS3.8)

=12

[X,Y ] (PS3.9)

Z3 =16

ÄX3 + 3X2Y + 3XY 2 + Y 3

ä− 1

6(X + Y )3

− 12

Å(X + Y )

12

(XY − Y X) +12

(XY − Y X)(X + Y )ã

(PS3.10)

=16

Ä2X2Y + 2XY 2 −XYX − Y X2 − Y XY − Y 2X

ä− 1

4

ÄX2Y −XYX + Y XY − Y 2X +XYX +XY 2 − Y X2 − Y XY

ä(PS3.11)

=112

ÄX2Y +XY 2 − 2XYX − 2Y XY + Y X2 + Y 2X

ä(PS3.12)

=112

(X[X,Y ] + [X,Y ]Y + [Y,X]X + Y [Y,X]) (PS3.13)

=112

([X, [X,Y ]] + [Y, [Y,X]]) (PS3.14)

24Z4 =ÄX4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4

ä− (Z1)4

− 4Ä(Z1)2Z2 + Z1Z2Z1 + Z2(Z1)2

ä− 12

ÄZ1Z3 + (Z2)2 + Z3Z1

ä(PS3.15)

= X4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4 − (Z1)4

− 4(Z1)2Z2 − 4Z1Z2Z1 − 4Z2(Z1)2 − 12Z1Z3 − 12(Z2)2 − 12Z3Z1 (PS3.16)

= X4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4 − (Z1)4

− 6(Z1)2Z2 + 2Z1[Z1, Z2] + 2[Z2, Z1]Z1 − 6Z2(Z1)2

− 12Z1Z3 − 12(Z2)2 − 12Z3Z1 (PS3.17)

146

= X4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4 − (X2 + 2XY + Y 2 − 2Z2)2

− 6(X2 + 2XY + Y 2 − 2Z2)Z2 − 6Z2(X2 + 2XY + Y 2 − 2Z2)

+ 2Z1[Z1, Z2] + 2[Z2, Z1]Z1 − 12Z1Z3 − 12(Z2)2 − 12Z3Z1 (PS3.18)

= X4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4 − (X2 + 2XY + Y 2)2

+ 2Z2(X2 + 2XY + Y 2) + 2(X2 + 2XY + Y 2)Z2 − 4(Z2)2

− 6(X2 + 2XY + Y 2)Z2 + 12(Z2)2 − 6Z2(X2 + 2XY + Y 2) + 12(Z2)2

+ 2Z1[Z1, Z2] + 2[Z2, Z1]Z1 − 12Z1Z3 − 12(Z2)2 − 12Z3Z1 (PS3.19)

= X4 + 4X3Y + 6X2Y 2 + 4XY 3 + Y 4 − (X2 + 2XY + Y 2)2

− 4(X2 + 2XY + Y 2)Z2 − 4Z2(X2 + 2XY + Y 2)

+ 2Z1[Z1, Z2] + 2[Z2, Z1]Z1 − 12Z1Z3 + 8(Z2)2 − 12Z3Z1 (PS3.20)

= 2[X2, XY ] + 2[XY, Y 2] + [X2, Y 2] + 4X[X,Y ]Y

− 2(X2 + 2XY + Y 2)[X,Y ]− 2[X,Y ](X2 + 2XY + Y 2)

+ Z1[Z1, [X,Y ]] + [[X,Y ], Z1]Z1 + 2[X,Y ]2

− Z1[X, [X,Y ]]− Z1[Y, [Y,X]]− [X, [X,Y ]]Z1 − [Y, [Y,X]]Z1 (PS3.21)

= 2X2[X,Y ] + 2X[X,Y ]X + 2[X,Y ]Y 2 + 2Y [X,Y ]Y + [X2, Y 2] + 4X[X,Y ]Y

− 2(X2 + 2XY + Y 2)[X,Y ]− 2[X,Y ](X2 + 2XY + Y 2)

+ 2Z1[Y, [X,Y ]] + 2[X, [Y,X]]Z1 + 2[X,Y ]2 (PS3.22)= 2X[X,Y ]Z1 + 2Z1[X,Y ]Y +XY [X,Y ] +X[X,Y ]Y + Y [X,Y ]X + [X,Y ]Y X

− 4XY [X,Y ]− 2Y 2[X,Y ]− 2[X,Y ]X2 − 4[X,Y ]XY

+ 2Z1[Y, [X,Y ]] + 2[X, [Y,X]]Z1 + 2[X,Y ]2 (PS3.23)= 2[X, [X,Y ]]X + 3X[[X,Y ], Y ] + 2[X, [X,Y ]]Y + 2Y [[X,Y ], Y ]− [[X,Y ], Y ]X

+ 2Z1[Y, [X,Y ]] + 2[X, [Y,X]]Z1 (PS3.24)= −X[Y, [X,Y ]] + [Y, [X,Y ]]X (PS3.25)= [[Y, [X,Y ]], X] (PS3.26)

Z4 = − 124

[X, [Y, [X,Y ]]] (PS3.27)

**

3. Let F (d) be the free Lie algebra on generatorsX1, . . . , Xd. It has a natural Nd grading in whichF (d)(k1,...,kd) is spanned by bracket monomials containing ki occurences of each generator Xi.Use the PBW theorem to prove the generating function identity∏

k

1

(1− tk11 . . . tkdd )dimF (d)(k1,...,kd)=

11− (t1 + · · ·+ td)

4. Words in the symbols X1, . . . , Xd form a monoid under concatentation, with identity theempty word. Define a primitive word to be a non-empty word that is not a power of a shorter

147

word. A primitive necklace is an equivalence class of primitive words under rotation. Use thegenerating function identity in Problem 3 to prove that the dimension of F (d)k1,...,kd is equalto the number of primitive necklaces in which each symbol Xi appears ki times.

5. A Lyndon word is a primitive word that is the lexicographically least representative of itsprimitive necklace.

(a) Prove that w is a Lyndon word if and only if w is lexicographically less than v for everyfactorization w = uv such that u and v are non-empty.

**Throughout, we use the notation that a < b if a and b are words and ais lexicographically less than b. We let |a| be the multigrading of a: if ouralphabet has d letters in it, then |a| is the vector of length d, whose kth entryis the number of times the kth letter appears in the word a. Multigradingshave an obvious partial order, and in general we will say that “a is longerthan b” if |a| > |b|. We never use sub- or super-scripts to denote powers,always writing aa for the concatenation of a with itself.

We remark, and will use repeatedly, the fact that if a < b and a is not a prefixof b, then ac < b for any word c. Moreover, if a < bc and b is not a prefix of a,then a < b. We remark also that a < ac for any suffix c; if b < c, then ab < acfor any a.

We let w = uv be Lyndon but v ≤ w, looking for a contradiction, and wechoose v to be the shortest such suffix of w. Then v 6= u since w is not apower, and v 6= w as u is nonempty. Of course, vu > w as w is Lyndon, i.e.uv < vu. By the previous remark, v must be a prefix of w. Hence either v isa prefix of u or u is a prefix of v. In the first case, w = vu′v, and in the secondcase w = uuv′. We treat the second case first. We have w = uuv′ > uv′, hencev′ < uv′ = v < w. But we insisted that v be the shortest suffix of w less than w,whence v′ cannot be less than w as it is strictly shorter. Thus, we must be inthe first case: w = vu′v. If vu′ < u′v then vvu′ < vu′v = w, and if u′v < vu′ thenu′vv < vu′v as vu′ and u′v have the same length so neither is a prefix of theother. The only option is that u = vu′ = u′v, and we remark that a word isonly fixed under a cyclic permutation if it is a power, say u = tt . . . t, in whichcase u′ and v are both necessarily powers of t, hence w is not primitive.

Conversely, if any proper suffix v of w (i.e. any word v so that w = uv forsome u) is greater than w, then by concatenating the corresponding prefixvu we get an even larger word, so w is Lyndon.

We observe that if a and b are Lyndons with a < b, then ab is Lyndon. Towit: If a = a1a2, then being Lyndon we have a < a2, and since a is longer, itis not a prefix. Hence ab < a2 < a2b. If a is not a prefix of b, then b > ab. Ifb = ab′, then b′ > b, and so b = ab′ > ab. If b = b1b2, then b2 > b > ab. Thereforeany suffix of ab is larger than ab, and so ab is Lyndon. **

148

(b) Prove that if w = uv is a Lyndon word of length > 1 and v is the longest properright factor of w which is itself a Lyndon word, then u is also a Lyndon word. Thisfactorization of w is called its right standard factorization.

**We write wR for the longest proper Lyndon suffix of w, and wL for thecorresponding prefix; thus w = wLwR is the right standard factorization of wif w is Lyndon. If w is not Lyndon, by (a), wR is the longest proper suffix ofw that is less than any shorter suffix. We remark that any word has a uniquefactorization into nonincreasing Lyndons: if w is not Lyndon, then either wLis Lyndon and at least wR, or (wL)RwR is not Lyndon, so (wL)R ≥ wR, and wecontinue factoring off longest Lyndon suffixes from the right. By inspection,if w is not Lyndon, then w > wR. Of course, by (a), if w is Lyndon, w < wR.

We wish to show that if w is Lyndon, then so is wL. Assume not; then(wL)R < wL < wR. Thus (wL)RwR is Lyndon, a suffix of w, and longer than wR,a contradiction.**

(c) To each Lyndon word w in symbols X1, . . . , Xd associate the bracket polynomial pw = Xi

if w = Xi has length 1, or, inductively, pw = [pu, pv], where w = uv is the right standardfactorization, if w has length > 1. Prove that the elements pw form a basis of F (d).

**We drop the p• notation, writing “a” for both the word a and, if a isLyndon, for the image of a in the free Lie algebra on d generators. Byconsidering the action when all the generators commute, we see that thebracket of Lyndon words preserves the multigrading (where we construct amultigrading of Lie-bracket monomials in the obvious way). By problem 4above, it suffices to show that the Lyndon words with a given multigradingspan the vector space generated by the Lie monomials at that grading. Byinduction on grading and by the antisymmetry of the bracket, it suffices toshow that if a and b are Lyndon and a < b then [a, b] is a sum of Lyndon words(necessarily of multigrading |a| + |b|). In this notation, the definition of themap from Lyndon words to bracket monomials is given by w = [wL, wR].

We show something slightly stronger to make the (strong transfinite) induc-tion go through: that if a < b, then [a, b] is a sum of Lyndon words lexicograph-ically less than b. Of course, ab < b as ab is Lyndon, and if (ab)L = a, then [a, b] =ab is Lyndon. Hence we need only consider the case that (ab)L 6= a. Assumethat aR < b. By the Jacobi identity, [a, b] = [[aL, aR], b] = [[aL, b], aR] + [aL, [aR, b]].By induction on grading, [aL, b] and [aR, b] are sums of Lyndon words less thanb. Let c1 and c2 be the largest Lyndon words that appear in the sums [aL, b]and [aR, b], respectively. Then max(c1, aR) and max(aL, c2) are both less thanb, so by induction each of [[aL, b], aR] and [aL, [aR, b]] is a sum of Lyndon wordsless than these maximums, and in particular less than b.

It suffices to show that if (ab)L 6= a, then aR < b. Since b is a Lyndon suffix ofab, we necessarily have that (ab)R is longer than b: a = (ab)La2 and (ab)R = a2b.

149

If aR is at least as long as a2 then aR ≤ a2 < a2b < b as aR and (ab)R = a2bare Lyndon. If aR is shorter than a2, then a2 is not Lyndon, aR is its longestLyndon suffix, and aR < a2, whence aR < a2b < b. This completes the proof.**

6. Prove that if q is a power of a prime, then the dimension of the subspace of total degreek1 + · · · + kq = n in F (q) is equal to the number of monic irreducible polynomials of degreen over the field with q elements.

7. This problem outlines an alternative proof of the PBW theorem.

**I didn’t attempt this problem. I will mention that the solution above to prob-lem 5 considers only whether the Lyndon monomials span their graded compo-nent, and so the proof could be be reproduced verbatim for part (b) below.**

(a) Let L(d) denote the Lie subalgebra of T (X1, . . . , Xd) generated by X1, . . . , Xd. Withoutusing the PBW theorem—in particular, without using F (d) = L(d)—show that the valuegiven for dimF (d)(k1,...,kd) by the generating function in Problem 3 is a lower bound fordimL(d)(k1,...,kd).

(b) Show directly that the Lyndon monomials in Problem 5(b) span F (d).

(c) Deduce from (a) and (b) that F (d) = L(d) and that the PBW theorem holds for F (d).

(d) Show that the PBW theorem for a Lie algebra g implies the PBW theorem for g/a,where a is a Lie ideal, and so deduce PBW for all nitely generate Lie algebras from (c).

(e) Show that the PBW theorem for arbitrarty Lie algebras reduces to the finitely generatedcase.

8. Let B(X,Y ) be the Baker-Campbell-Hausdorff series, i.e., eB(X,Y ) = eXeY in noncommutingvariables X, Y . Let F (X,Y ) be its linear term in Y , that is, B(X, sY ) = X + sF (X,Y ) +O(s2).

(a) Show that F (X,Y ) is characterized by the identity

∑k,l≥0

Xk F (X,Y )X l

(k + l + 1)!= eXY. (PS3.28)

**We let B(X, sY ) = X + sF (X,Y ) + O(s2) and exponentiate, using eB(X,sY ) =

150

eXesY :

eXesY = eXÄ1 + sY +O(s2)

ä(PS3.29)

eB(X,sY ) = eX+sF (X,Y )+O(s2) (PS3.30)

=∞∑n=0

1n!

(X + sF (X,Y ))n +O(s2) (PS3.31)

=∞∑n=0

1n!

(Xn +

n−1∑k=0

XksF (X,Y )Xn−k−1 +O(s2)

)+O(s2) (PS3.32)

Comparing terms linear in s in (PS3.29) and (PS3.32) we recover exactly(PS3.28). **

(b) Let λ, ρ denote the operators of left and right multiplication by X, and let f be theseries in two commuting variables such that F (X,Y ) = f(λ, ρ)(Y ). Show that

f(λ, ρ) =λ− ρ

1− eρ−λ

**We observe that F (X,Y ) is of homogeneous degree 1 in Y , hence it makessense to define f as above. In this notation, part (a) asserts that

∑k,l≥0

λkρl

(k + l + 1)!f(λ, ρ) = eλ (PS3.33)

where we interpret everything in terms of formal power series.

Let g(λ, ρ) be the polynomial multiplying f(λ, ρ) in the right-hand-side of(PS3.33). We can find a closed-form expression for g:

g(λ, ρ) =∑k,l≥0

λkρl

(k + l + 1)!(PS3.34)

g(λ, ρ) (λ− ρ) =∑k,l≥0

λk+1ρl − λkρl+1

(k + l + 1)!(PS3.35)

=∑k≥1,l≥0

λkρl

(k + l)!−∑k≥0,l≥1

λkρl

(k + l)!(PS3.36)

=∑k≥1,l=0

λkρl

(k + l)!−∑k=0,l≥1

λkρl

(k + l)!(PS3.37)

=∑k≥1

λk

k!−∑l≥1

ρl

l!(PS3.38)

= (eλ − 1)− (eρ − 1) (PS3.39)

151

Thus g(λ, ρ) = (eλ − eρ)/(λ− ρ), and

f(λ, ρ) =eλ

g(λ, ρ)=

λ− ρ1− eρ−λ

(PS3.40)

**

(c) Deduce that

F (X,Y ) =adX

1− e− adX(Y ).

**We remark that (λ− ρ)Y = (adX)Y for any Y , since λ is left-multiplicationby X and ρ is right-multiplication by X (these operations commute). Theresult follows from the definition that F (X,Y ) = f(λ, ρ)(Y ) and (PS3.33),remembering as always that every expression should be understood as aformal power series.**

9. Let G be a Lie group, g = Lie(G), 0 ∈ U ′ ⊆ U ⊆ g and e ∈ V ′ ⊆ V ⊆ G open neighborhoodssuch that exp is an isomorphism of U onto V , exp(U ′) = V ′, and V ′V ′ ⊆ V . Define β :U ′ × U ′ → U by β(X,Y ) = log(exp(X) exp(Y )), where log : V → U is the inverse of exp.

(a) Show that β(X, (s+ t)Y ) = β(β(X, tY ), sY ) whenever all arguments are in U ′.

**When all arguments are in U ′, we can exponentiate and compare in V ′.Then the right-hand-side is exp(X) exp((s+ t)Y ), whereas the left-hand-side isexp(β(X, tY )) expY = exp(X) exp(tY ) exp(sY ), clearly equal.**

(b) Show that the series (adX)/(1 − e− adX), regarded as a formal power series in thecoordinates of X with coefficients in the space of linear endomorphisms of g, convergesfor all X in a neighborhood of 0 in g.

**We saw above that as formal power series,

adX1− e− adX

Y =d

dt

∣∣∣∣t=0

B(X, tY ) (PS3.41)

But B(X, tY ) converges to β(X, tY ) for small t and X in a neighborhood of0, so the right-hand-side of (PS3.41) converges to d

dt

∣∣∣t=0

β(X, tY ) for X in aneighborhood of 0.

Oh, but a later problem, building on this one, would like to conclude withan alternate proof that B(X,Y ) → β(X,Y ) in a neighborhood. So we shouldshow the above convergence directly.

152

We abbreviate x = adX. Then

x

1− e−x=

−x∑n≥1

(−x)n

n!

(PS3.42)

=1

1− x∑n≥0

(−x)n

(n+2)!

(PS3.43)

When x is very small, then∑n≥0

(−x)n

(n+2)! is close to 1, and so the geometric seriesin (PS3.43) converges. If g is finite-dimensional, then adX is small in matrixnorm provided that X is in a neighborhood of the origin. If g is infinite-dimensional, we should demand that whatever topology it has, ad : g→ gl(g)be linear. **

(c) Show that on some neighborhood of 0 in g, β(X, tY ) is the solution of the initial valueproblem

β(X, 0) = X (PS3.44)d

dtβ(X, tY ) = F (β(X, tY ), Y ), (PS3.45)

where F (X,Y ) =Ä(adX)/(1− e− adX)

ä(Y ).

**The proposed solution β(X,Y ) = log(expX expY ) converges on a neighbor-hood, and so on this neighborhood β(X, tY ) converges when |t| ≤ 1. It clearlysatisfies (PS3.44) when X is in this neighborhood. We differentiate, usingpart (a):

d

dtβ(X, tY ) = lim

s→0

β(X, (t+ s)Y )− β(X, tY )s

(PS3.46)

= lims→0

β(β(X, tY ), sY )− β(β(X, tY ), 0)s

(PS3.47)

=d

dsβ(β(X, tY ), sY )

∣∣∣∣s=0

(PS3.48)

So it suffices to show that

d

dsβ(X, sY )

∣∣∣∣s=0

=adX

1− e− adX(Y ) (PS3.49)

**

(d) Show that the Baker-Campbell-Hausdorff series B(X,Y ) also satises the identity in part(a), as an identity of formal power series, and deduce that it is the formal power seriessolution to the IVP in part (c), when F (X,Y ) is regarded as a formal series.

153

**We treat everything as formal power series in noncommuting variables.The BCH series B(X,Y ) is defined by B(X,Y ) = log

ÄeXeY

ä. Then

B (X, (t+ s)Y ) = logÄeXe(t+s)Y

ä= log

ÄeXetY esY

ä=

= logÄeB(X,tY )esY

ä= B(B(X, tY ), sY ) (PS3.50)

Of course, B(X,Y ) satisfies (PS3.44). To check (PS3.45), we differentiate:

d

dtB(X, tY ) = lim

s→0

B(X, (t+ s)Y )−B(X, tY )s

(PS3.51)

= lims→0

B(B(X, tY ), sY )−B(B(X, tY ), 0)s

(PS3.52)

= F (B(X, tY ), Y ) (PS3.53)

We use problem 8, which shows that as formal power series ddsB(X, sY )

∣∣∣s=0

=F (X,Y ). **

(e) Deduce from the above an alternative proof that B(X,Y ) is given as the sum of a seriesof Lie bracket polynomials in X and Y , and that it converges to β(X,Y ) when evaluatedon a suitable neighborhood of 0 in g.

**Since β(X,Y ) and B(X,Y ) solve the same initial value problem, in particularthey must have all the same derivatives at the origin. I will skip the argumentthat β is analytic; it is if exp and the Lie group multiplication are. In anearlier problem set, I claimed to prove that many things were analytic, butin fact only proved that their power-series expansions have infinite radius ofconvergence. Analyticity in those cases follows by repeating my argumentin a neighborhood of the origin (using such facts as that for any analyticfunction, there is a neighborhood of the origin and a positive real number aso that the nth derivative of the function is less in absolute value than n!an).But here I don’t know how to show analyticity particularly well.

In any case, since B(X, tY ) solve the IVP in equations (PS3.44-PS3.45), wecan compute each term. Indeed,Åd

dt

ãnB(X, tY )

∣∣∣∣t=0

=Åd

dt

ãn−1

F (B(X, tY ), Y )

∣∣∣∣∣t=0

(PS3.54)

= (derivatives of F )(B(X, tY ), Y ) · (derivatives of B(X, tY ))∣∣∣t=0

(PS3.55)

By induction, the derivatives of B, which are of degree at most n − 1, arebracket polynomials of X and tY . The derivatives of F act on the derivativesof B by brackets. **

154

(f) Use part (c) to calculate explicitly the terms of B(X,Y ) of degree 2 in Y .

**For example, we compute the terms of degree 2 in Y in B(X,Y ). We letz = adβ(X, tY ). Then, since ad : g → gl(g) is linear, we have z

def= ddt

∣∣∣t=0

z =adF (X,Y ) is the adjoint action of a bracket polynomial.Åd

dt

ã2

B(X, tY )

∣∣∣∣∣t=0

=d

dtF (B(X, tY ), Y )

∣∣∣∣t=0

=d

dt

z

1− e−z∣∣∣∣t=0

Y

=Ä1− e−z

ä−1ÅÄ

1− e−zäz − z d

dt

Ä1− e−z

ä∣∣∣∣t=0

ã Ä1− e−z

ä−1(Y )

=Ä1− e−z

ä−1

Ñz − e−z z − z

∑k,l≥0

(−z)kz(−z)l

(k + l + 1)!

éÄ1− e−z

ä−1(Y )

=Ä1− e−z

ä−1

Ñz −

∑k≥0

(−z)kzk!

+∑k,l≥0

(−z)k+1z(−z)l

(k + l + 1)!

éÄ1− e−z

ä−1(Y )

=Ä1− e−z

ä−1

Ñz +

∑k,l≥1

(−z)kz(−z)l

(k + l)!

éÄ1− e−z

ä−1(Y )

This is a bracket polynomial in X and Y , as both z and z act as the adjointaction of bracket polynomials of X and Y on everything to their right. **

10. (a) Show that the Lie algebra so(3,C) is isomorphic to sl(2,C).

(b) Construct a Lie group homomorphism SL(2,C) → SO(3,C) which realizes the isomor-phism of Lie algebras in part (a), and calculate its kernel.

**We do the second part first, the first part then following by general nonsense.

SL(2) acts on its defining representation 2 = C2, and hence on the tensor squareC4 = 2 ⊗ 2. If {e1, e2} are an ordered basis of C4, then {e11, e12, e21, e22} are an

ordered basis of 2 ⊗ 2, where eij = ei ⊗ ej. Given x =ña bc d

ô∈ SL(2) (so in

particular ad− bc = 1), we have the action of x on C4 given by the matrix

x 7→

a2 ab ba b2

ac ad bc bdca cb da dbc2 cd dc d2

(PS3.56)

Since ad− bc = detx = 1, we see that x acts trivially on e12 − e21, and asña bc d

ô7→

a2√

2ab b2√2ac ad+ bc

√2bd

c2√

2cd d2

(PS3.57)

155

on the subspace given by the ordered basis {e11, (e21 + e12)/√

2, e22}. Hence 2⊗ 2 =1⊕3, where 1 is the trivial representation of SL(2) and 3 is the three-dimensionalrepresentation given by (PS3.57). But the matrix in (PS3.57) preserves the dotproduct given in matrix form by 1

−11

(PS3.58)

as can be seen by explicit calculation, using the fact that ad − bc = 1 and hencea2d2−2abcd+b2c2 = 1. Thus, at least over C where all (nondegenerate) dot productsare the same, we see that SL(2) is acting on C3 by matrices in SO(3), and this givesthe map. Of course, the kernel is exactly those matrices in SL(2) with b = c = 0and a2 = d2 = ad = 1; these are exactly ±1 ∈ SL(2).

Since the kernel is discrete, corresponding (differential) Lie algebra homomor-phism sl(2)→ so(3) is an injection, and by dimension count necessarily an isomor-phism.

We remark that we could have started with a different(ly presented) four-dimensionalrepresentation. There is a nontrivial automorphism of SL(2) given by sendingeach matrix to its transpose inverse. By composing with this automorphism,SL(2) acts on C2 in the “dual representation” 2, where now we write the indicesraised: the basis of 2 is {e1, e2}, and the basis of 2 ⊗ 2 is {eji} where eji = ei ⊗ ej.Then the action SL(2) y 2⊗ 2 is the adjoint action on gl(2). Of course, 2 ∼= 2 viae1 = e2, e2 = −e1, and the splitting 2 ⊗ 2 = 1 ⊕ 3 splits gl(2) into the span of theidentity e1

1 + e22 and the adjoint representation of SL(2) on sl(2), now spanned by

{e21, e

12, (e

11 + e2

2)/√

2}. Which is all to say that 3 is the adjoint representation ofSL(2). **

11. (a) Show that the Lie algebra so(4,C) is isomorphic to sl(2,C)× sl(2,C).

(b) Construct a Lie group homomorphism SL(2,C) × SL(2,C) → SO(4,C) which realizesthe isomorphism of Lie algebras in part (a), and calculate its kernel.

**We continue the explicit calculations at the level of groups. SO(4) is generatedby the rotations that fix a plane. We let xij(t), where t is a complex parameter,

be the matrix that rotates the ei, ej-plane byñ

cos t − sin tsin t cos t

ôand fixes the perpen-

dicular plane, where {e1, . . . , e4} is our basis of the defining representation. Forexample:

x12(t) =

cos t − sin tsin t cos t

11

(PS3.59)

156

Then xji(t) = xij(t)−1 = xij(−t), so we can assume i < j, and xij and xkl commute if{i, j} ∩ {k, l} = ∅.

We will actually want a different generating set. We let a±(t) = x12(t)x34(±t),b±(t) = x13(t)x24(∓t), and c±(t) = x14(t)x23(±t). These six functions also generateSO(4) as t ranges. It’s immediate that a+(t) and a−(s) commute, and similarlyfor b and c. But, in fact, one can easily calculate that a+(t) and b−(s) commute,and indeed the functions {a+, b+, c+} and {a−, b−, c−} each generate a subgroup ofSO(4), and the two subgroups commute. We call the subgroups G+ and G−. Thenthe product G+G− = SO(4), since x12(t) = a+(t/2)a−(t/2), etc. Since G+ and G−commute, any element of their intersection must commute with all of G+ and allof G−, hence with a generating set of SO(4), and thus is central. But the centerof SO(4) is ±1, and indeed −1 ∈ G+ ∩G−. Hence SO(4) ∼= G+ ×G−/H, where H isthe two-element subgroup whose nontrivial element is (−1,−1) ∈ G+ ×G−.

Consider the vectors f− = e1 − ie2 and g+ = e3 + ie4. One easily has a+(t) : f− 7→cos(t)f− + i sin(t)f− = eitf−, and similarly a+(t) : g+ 7→ e−itg+. Moreover, b+(t) : f− 7→cos(t)f− + sin(t)g+ and so on. G+ acts on the vector space spanned by {f−, g+}:

a+(t) =ñeit 00 e−it

ô, b+(t) =

ñcos(t) − sin(t)sin(t) cos(t)

ô, c+(t) =

ñcos(t) −i sin(t)−i sin(t) cos(t)

ô(PS3.60)

I.e. G+ is acting as SL(2) on the span 〈f−, g+〉: we have a map G+ → SL(2). G+

also acts as SL(2) on the perpendicular space 〈f+, g−〉, where f+ = e1 + ie2 andg− = e3 − ie4. The matrices above are the same except for the obvious complexconjugation. G− acts by similar matrices on 〈f−, g−〉 and on 〈f+, g+〉.

The maps G± → SL(2) given in (PS3.60) are clearly onto, since the matrices listedgenerate SL(2). But they are also necessarily one-to-one: indeed, let x ∈ G+ act asthe identity on f− and g+. Then the action of x on f+ and g− is given by writingx as a product of as, bs, and cs, (each of a different value of t) and switchingthe is the −is in the cs, thus, if x fixes f− and g+, then it fixes f+ and g−. But{f±, g±} are a basis, so x = 1 and G± → SL(2) are isomorphisms. This givesSO(4) ∼= SL(2)× SL(2)/H. **

12. Show that every closed subgroup H of a Lie group G is a Lie subgroup, so that the inclusionH ↪→ G is a closed immersion.

**By the first proposition on the 12th of September, to show that H ↪→ G isan immersion it suffices to find for each point p ∈ H a chart U 3 p with localcoordinates ξi in G so that H ∩ U = {q ∈ U s.t. ξn+1(q) = · · · = ξm(q) = 0}, whereG is of dimension m. By multiplying by q−1, it suffices to do this at the origine ∈ H ⊆ G.

But any Lie group G has a neighborhood U around the origin and a diffeomor-phism φ : U ∼→ V where V is a neighborhood of 0 in Lie(G). We restrict to a compact

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neighborhood so that everything is uniform. Well, φ(U∩H) is almost a sub-partial-group of the additive partial-group Lie(G): φ(φ−1(a)φ−1(b)) = a + b + O(a, b)2, andby uniform continuity this big-O bound in uniform. Thus, we can shrink thesize of V so that relative to a + b we can ignore the O(a, b)2 error. But the onlysubgroups of Rm are of the form Rn × discrete. The Rn part we’re happy with —indeed, by definition unpacking, this is just Lie(H) — and the rest we can avoidby restricting our neighborhood. **

13. Let G be a Lie group and H a closed subgroup. Show that G/H has a unique manifoldstructure such that the action of G on it is smooth [analytic, holomorphic].

**The action of G on itself if smooth, and so the cosets of H are also im-mersed manifolds. Let U, ξ be a chart at the identity in G so that H ∩ U ={q ∈ U s.t. ξn+1(q) = · · · = ξm(q) = 0}. Cosets gH of H cannot intersect each other,and so the various gH ∩ Us have no choice but to foliate U . We restrict U sothat the gHs never get a chance to “fold back” on themselves: i.e. so that foreach coset gH with nonempty intersection with U , it has a unique point for anygiven (small) values of the coordinates ξ1, . . . , xn. (We can do this by restrictingU to a small compact neighborhood.) Let p(gH) be the point in gH ∩ U so thatξ1(p) = · · · = ξn(p) = 0, and define ζi(gH) = ξn+i(p(gH)) for 1 ≤ i ≤ m − n. Thisdefines a chart on the cosets, for which the G action is smooth.**

14. Show that the intersection of two Lie subgroups H1, H2 of a Lie group G can be given acanonical structure of Lie subgroup so that its Lie algebra is Lie(H1) ∩ Lie(H2) ⊆ Lie(G).

**The first theorem on the first of October says that any Lie subalgebra of Lie(G) isthe algebra of a unique connected subgroup of G. In particular, Lie(H1)∩Lie(H2) isa subalgebra of Lie(G), so is the Lie algebra of a unique connected subgroup J , butit is also a subalgebra of each Hi, and so J is in fact a connected subgroup of H1∩H2.Let K be the set of cosets of J in H1 ∩H2, which we give the discrete topology;then H1 ∩ H2 can be given a manifold topology as J ×K. It’s easy to constructsmooth maps out of discrete sets, and in particular the map H1∩H2 = J ×K → Hi

is s/a/h.**

15. Find the dimension of the closed linear group SO(p, q,R) ⊆ SL(p+q,R) consisting of elementswhich preserve a non-degenerate symmetric bilinear form on Rp+q of signature (p, q). Whenis this group connected?

**Of course, the complexification SO(p, q,C) = C⊗R SO(p, q,R) cannot see the signof the non-degenerate form: SO(p, q,C) = SO(p + q,C). But dimR SO(p, q,R) =dimC SO(p, q,C) = dimC SO(p+ q,C) = dimR SO(p+ q,R) =

(p+q2

)= (p+ q)(p+ q − 1)/2

We can compute this directly, of course. Let gmn be a symmetric bilinear formof signature (p, q). Then Xj

i ∈ SO(p, q,R) if an only if Xmi g

ijXnj = gmn. Since g is

symmetric, this condition consists of(p+q+1

2

)equations, each of which cuts one

dimension: thus dimSO(p, q,R) = (p+ q)2 −(p+q+1

2

)=(p+q

2

). Working infinitesimally,

158

we differentiate Xmi g

ijXnj = gmn to get the requirement on the Lie algebra that

Xmi g

in+gmjXnj = 0. Working with matrices, this corresponds to the condition that

X + XT = 0, where now the transpose switches the sign of the off-diagonal p × qblocks. Same number of equations, same size of algebra.

We remark that SO(p, q) = SO(q, p), since if a matrix preserves a form, then itpreserves the negation of the form.

In any case, we know from a previous assignment that SO(p, 0) is connected. Thecolumns of any matrix X ∈ SO(p, q) comprise a basis of Rp+q the first p vectors ofwhich have length +1 and the last q have length −1 (with the added condition thatdifferent basis vectors are pairwise orthogonal), and a path of matrices consists ofa path of such bases. Thus SO(p, 1) is not connected: the reflection in the first andlast coordinates has determinant +1, but there is no path among the “time-like”vectors of length −1 between the “forward-pointing” and “backwards-pointing”sections of the cone.

In fact, it’s easy enough to see that SO(p, q) is never connected when pq > 0. Let

X =ñA B

C D

ôbe a matrix in SO(p, q), where A is p×p, D is q× q, etc. We take the

dot-product of signature (p, q) to be ~x ·~y = x1y1 + · · ·+xpyp−xp+1yp+1−· · ·−xp+qyp+q;if we think of ~x = ~x+ ⊕ ~x−, where ~x+ is of length p and ~x− is of length q, then~x · ~y = ~x+ · ~y+ − ~x− · ~x−, where now the dot-product is the usual Euclidean one.Then we can state the condition that X ∈ SO(p, q) in terms of the dot product.For each i = 1, . . . , p and j = 1, . . . , q, let Ai, Bj, Ci, and Dj be the i or jth columnof A, . . . ,D. X ∈ SO(p, q) if and only if Ai ·Ai−Ci ·Ci = 1 = Di ·Di−Bi ·Bi for each i,Ai ·Aj −Ci ·Cj = 0 = Bi ·Bj −Di ·Dj for i 6= j, and Ai ·Bj −Ci ·Dj for any pair (i, j).

We claim that detA is not zero. Indeed, If detA = 0, then the columns of A span aspace of dimension at most p− 1 in Rp, and there is a rotation in SO(p) ↪→ SO(p, q)bringing this space into the span of the first p − 1 standard basis vectors. Thus,up to a rotation, A has a column Ap of all 0s. But then the pth column of thisrotated X cannot have length 1 (indeed, it cannot have positive length), and soX 6∈ SO(p, q).

Thus along any path in SO(p, q) the upper-left p× p minor does not change sign.Hence there is no path from the identity matrix to the reflection in the first andlast coordinates.

We remark that SO(p, q) by the rotations SO(p) × SO(q) ↪→ SO(p, q) along with

159

boosts of the form

cosh t sinh t1

. . .1

sinh t cosh t1

. . .1

In fact, it suffices to take just one (one-parameter family of) boosts, as the restare conjugate by rotations. By inspecting the Lie algebra so(p, q), we see that thecommutator of infinitesimal boosts is an infinitesimal rotation, and so the boostsgenerate a dense subgroup of SO(p, q). **

16. Show that the kernel of a Lie group homomorphism G→ H is a closed subgroup of G whoseLie algebra is equal to the kernel of the induced map Lie(G)→ Lie(H).

**The kernel of a homomorphism is always a subgroup, and being the inverseimage of the closed set {e} ⊆ H, the kernel is necessarily closed in G.

We let φ : G → H be a Lie group homomorphism, whence dφ : Lie(G) → Lie(H)is the map of tangent spaces. If x(t) is a path in kerφ with x(0) = e ∈ G, thenφ(x(t)) = e ∈ G for each t. Hence dφ · x′(0) = d

dt

∣∣∣t=0

φ(x(t)) = ddt

∣∣∣t=0

e = 0, and sox′(0) ∈ ker dφ. Thus Lie(kerφ) ⊆ ker dφ. Conversely, let x(t) with x(0) = e be anypath in G such that dφ · x′(0) = 0. Then by working in a local chart of e we seethat x(t) = ξ(t) + o(t) where φ(ξ(t)) = e ∈ H and ξ(0) = e ∈ G. Thus x is tangent toa path in kerφ, and ker dφ ⊆ Lie(kerφ). **

17. Show that if H is a normal Lie subgroup of G, then Lie(H) is a Lie ideal in Lie(G).

**Of course, we have already shown that Lie(H) is naturally a subalgebra ofLie(G). If H is normal in G, then ghg−1 ∈ H for g ∈ G and h ∈ H. The conjugationmap G × H → H : (g, h) 7→ ghg−1 is a continuous (s/a/h) group homomorphismand in particular fixes the identity e (indeed, it’s the identity on {e} × H andsends G× {e} → {e}). Differentiating in h provides an action of G on Lie(H), anddifferentiating this action in G provides a Lie action Lie(G) y Lie(H). But thisaction is just the adjoint action, and Lie(H) is an ideal of Lie(G). **


1. Classify the 3-dimensional Lie algebras g over an algebraically closed field K of characteristiczero by showing that if g is not a direct product of smaller Lie algebras, then either

160

• g ∼= sl(2,K),

• g is isomorphic to the nilpotent Heisenberg Lie algebra h with basis X,Y, Z such that Zis central and [X,Y ] = Z, or

• g is isomorphic to a solvable algebra s which is the semidirect product of the abelianalgebra K2 by an invertible derivation. In particular s has basis X,Y, Z such that[Y,Z] = 0, and adX acts on KY +KZ by an invertible matrix, which, up to change of

basis in KY +KZ and rescaling X, can be taken to be eitherñ

1 10 1

ôorñλ 00 1

ôfor

some nonzero λ ∈ K.

**We assume first that g is generated by two elements X and Y . Then Z = [X,Y ] 6∈KX +KY , and {X,Y, Z} is a basis of g. We remark that

[aX + bY, cX + dY ] = detÇa bc d

å[X,Y ] (PS4.1)

Hence the operation πXY adZ =“bracket with Z, then project along Z to theX,Y -plane” depends only on the plane and a choice of volume form. Since K isalgebraically closed, we can find a basis with determinant 1 relative to X,Y suchthat πXY adZ is diagonalized, unless it has a repeated eigenvalue.

We treat first the diagonalizable case, whence we can assume that [Z,X] = aX+αZand [Z, Y ] = bY + βZ. We use the Jacobi identity:

0 = [[Z,X], Y ] + [[X,Y ], Z] + [[Y, Z], X] (PS4.2)= [aX + αZ, Y ] + [Z,Z] + [bY + βZ,X] (PS4.3)= aZ + α(bY + βZ)− bZ + β(aX + αZ) (PS4.4)= aβX + bαY + (a− b+ 2αβ)Z (PS4.5)

We have a few possibilities:

• If α = β = 0, then a = b. If a = b = 0 then Z is central and g is Heisenberg.Otherwise, a = b 6= 0, and [Z,X] = aX, [Z, Y ] = aY . If we change X to 1

aX,then Z becomes 1

aZ, and [ 1aZ,

1aX] = a

a2X = 1aX and [ 1

aZ, Y ] = aaY = Y . Hence

we may assume that [Z,X] = X and [Z, Y ] = Y , and g ∼= sl(2,K).

• If α 6= 0 then b = 0 and a + 2αβ = 0. If β 6= 0 then a = 0 and −b + 2αβ = 0.Hence we cannot have α, β both non-zero unless we are in characteristic 2.Assume without loss of generality that α 6= 0 and β = 0. Then a + 0 = 0and so a = b = 0. Hence our algebra is generated by [X,Y ] = Z, [Z, Y ] = 0,and [Z,X] = αZ. Since [− 1

αX,Y ] = − 1αZ and [− 1

αZ,−1αX] = α

α2Z = −(− 1αZ), we

rescale X to − 1αX and achieve the algebra presented by

[X,Y ] = Z, [Y,Z] = 0, and [X,Z] = Z (PS4.6)

161

Then we can change basis Y 7→ Y − Z, and present the algebra by

[X,Y ] = 0, [Y,Z] = 0, and [X,Z] = Z (PS4.7)

But now Y is central, and g is a direct product of an abelian one-dimensionalalgebra with the nonabelian two-dimensional Lie algebra.

This takes care of the diagonalizable case.

In the non-diagonal case, we can still find one eigenvector of πXY adZ, and hence[Z,X] = aX + bY + αZ and [Z, Y ] = aY + βZ, with b 6= 0. Then the Jacobi identitygives

−aβX + (aα− bβ)Y + 2aZ = 0 (PS4.8)

Hence a = 0 = bβ. But b 6= 0, so β = 0, and our algebra is

[Y,Z] = 0, [X,Y ] = Z, and [X,Z] = −bY + αZ (PS4.9)

Thus we have an abelian Lie algebra spanned by Y,Z, on which adX acts by the

matrixÇ

0 −b1 −α

å. Since b 6= 0, this matrix is invertible, and so the eigenvalues are

not 0. We can divide X by one eigenvalue and assure that adX y span{Y,Z} has 1as an eigenvalue. Since K is algebraically closed, we can find a basis of span{Y, Z}that either diagonalizes adX or, if adX has a repeated eigenvalue (necessarily 1,

since we have 1 as one of the eigenvalues), then perhaps adX =Ç

1 10 1

åin some

basis.

Our last remark is on the initial assumption that g could be generated by only twoelements. Otherwise, for any X,Y ∈ g, [X,Y ] ∈ span{X,Y }. Let [X,Y ] = aX + bYand [X,Z] = cX + dZ. Then somethingX + multiple(yY + zZ) = [X, yY + zZ] =somethingX + byY + dzZ, so b = d, i.e. b depends only on X. By the same token, adepends only on Y , and there is some (necessarily linear) function f : g→ K suchthat [X,Y ] = −f(Y )X + f(X)Y for all X,Y ∈ g. As a linear functional from three-dimensional space to K, f must have a(n at least) two-dimensional kernel, whichis thus an abelian subalgebra. Let X be a vector not in this subalgebra. ThenadX acts by the scalar f(X). If f(X) = 0 then the algebra is abelian. Otherwisewe rescale X so that f(X) = 1. This gives one last algebra of the third type. **

1’. Following problems 28–35 in Knapp, Chapter I, classify the 3-dimensional Lie algebras overK when char(K) = 0 but K is not necessarily algebraically closed.

2. (a) Show that the Heisenberg Lie algebra h in Problem 1 has the property that Z acts nilpo-tently in every finite-dimensional module, and as zero in every simple finite-dimensionalmodule.

**Let h y V be a finite-dimensional h-module, and write X,Y, Z for theirimages in h→ gl(V ). Let p(x) =

∑pkx

k be the minimal polynomial of the action

162

of X on V , i.e. the lowest-degree polynomial such that 0 = p(X). It must existsince V is finite-dimensional. Then 0 = [p(X), Y ] = Z p′(X) since Z = [X,Y ]is central. Since p was minimal, p′(X) 6= 0, hence Z cannot be invertible.(If K has non-zero characteristic, then p′(x) might be the identically-zeropolynomial, and the argument fails here.) Thus Z has 0 as an eigenvalue.Since Z is central, any Z-eigenspace is a submodule of V . Thus V is simpleonly if Z acts as 0. More generally, ZV is a submodule of codimension atleast 1 in V on which h acts; by induction on dimension, ZdimZV = 0 on ZV ,so ZdimV = 0 on V and Z is nilpotent. **

(b) Construct a simple infinite-dimensional h-module in which Z acts as a non-zero scalar.[Hint: take X and Y to be the operators d/dt and t on K[t].]

**If p(t) is a polynomial in K[t], then by the product rule [ ddt , t]p(t) = ddt(t p(t))−

t ddtp(t) = p(t) + t ddtp(t)− tddtp(t) = p(t), so [ ddt , t] = 1 is central. Hence h ↪→ gl(K[t])

with X 7→ ddt , Y 7→ t, and Z 7→ 1. It’s clear that this is a simple module: if

p(t) is of degree n and K has characteristic 0, then p(n)(t) = Xnt is a non-zeroconstant polynomial, and any submodule containing p(t) contains K ⊆ K[t].But Y m1 = tm, so any submodule containing K contains all of K[t].

If K is not of characteristic 0, then K[t] is not simple; if n is a multiple ofthe characteristic of K, then the ideal generated by the tn is a submoduleof h y K[t]. But in this case K[t]/(tn) is a finite-dimensional module of h onwhich Z acts by multiplication by 1. **

3. Construct a simple 2-dimensional module for the Heisenberg algebra h over any field K ofcharacteristic 2. In particular, if K = K, this gives a counterexample to Lie’s theorem innon-zero characteristic.

**K[t]/t2, from above. Explicitly, we have a basis a, b, and X annihilates a andsends b to a, Y annihilates b and sends a to b, and then [X,Y ] = XY − Y X : a 7→(XY −Y X)a = Xb−Y 0 = a and b 7→ (XY −Y X)b = 0−Y a = −b = b; hence [X,Y ] actsas the identity, using the fact that we’re in characteristic 2, and is thus central.**

4. Let g be a finite-dimensional Lie algebra over K.

(a) Show that the intersection n of the kernels of all finite-dimensional simple g-modules canbe characterized as the largest ideal of g which acts nilpotently in every finite-dimensionalg-module. It is called the nilradical of g.

**If a is an ideal of g and V is a g-module, then aV is a sub-g-module; if a

acts nilpotently, then aV is a strict sub-module, and so if V is simple, thena must act as 0 on V . Hence, any ideal which acts nilpotently on everyfinite-dimensional module in particular acts as 0 on every finite-dimensionalsimple, and so is contained in n.

Conversely, let V be a finite-dimensional g-module and W a simple submod-

163

ule. Then n annihilates W and so nV is of lower dimension than V . Byinduction on dimension, ndim nV nV = 0, and so ndimV V = 0 and n acts nilpo-tently on every g-module. **

(b) Show that the nilradical of g is contained in g′ ∩ rad(g).

**g/g′ is an abelian Lie algebra on which g acts as 0; splitting g/g′ as a directsum of one-dimensionals, we can realize g′ as an intersection of kernels offinite-dimensional simples, and so g′ ⊇ n.

Then again, we’ve shown that every algebra has a faithful representation,and n acts nilpotently on this representation, thus n is nilpotent, and hencesolvable by a proposition from the 8th of October. Therefore n ⊆ rad g, sinceby definition rad g is the largest solvable ideal of g. **

(c) Let h ⊆ g be a subalgebra and V a g-module. Given a linear functional λ : h → K,define the associated weight space to be Vλ = {v ∈ V : Hv = λ(H)v for all H ∈ h}.Assuming char(K) = 0, adapt the proof of Lie’s theorem to show that if h is an idealand V is finite-dimensional, then Vλ is a g-submodule of V .

**We know that h y Vλ diagonally. Let X ∈ g, v ∈ Vλ, and H ∈ h. Then

HXv = XHv + [H,X]v (PS4.10)= Xλ(H)v + λ([H,X])v (PS4.11)

Thus E = XVλ+Vλ is a generalized eigenspace of H with eigenvalue λ (if Xv ∈Vλ, then HXv = λ(H)Xv already, and λ([H,X]) = 0). Thus trE H = dim(E)λ(H).By cyclicity of the trace, trE [H,X] = 0, but [H,X] ∈ h, so λ([H,X]) = 0 if Kis of characteristic 0. Thus HXv = λ(H)Xv, and X : Vλ → Vλ. So Vλ is ag-eigenspace. **

(d) Show that if char(K) = 0 then the nilradical of g is equal to g′ ∩ rad(g). [Hint: assumewithout loss of generality that K = K and obtain from Lie’s theorem that any finite-dimensional simple g-module V has a non-zero weight space for some weight λ on g′ ∩rad(g). Then use (c) to deduce that λ = 0 if V is simple.]

**Let g′ ∩ rad g = h. We have n ⊆ h. It suffices to show that h acts nilpotentlyon every finite-dimensional g-module, or equivalently as 0 on every simpleg-module.

Let V be a finite-dimensional g-module, hence it is a finite-dimensional h-module by restricting the action. But h is a submodule of the solvable rad g,and so by Lie’s theorem V has a one-dimensional h-submodule. On thissubmodule, h must act as scalars: if H ∈ h, then H acts on the submoduleby λ(H) for some λ : h→ K. Thus for this λ the weight-space Vλ is non-zero,since it contains this submodule.

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Thence (c) allows that Vλ is a non-zero submodule of V . If V is simple, thenVλ = V , and h acts diagonally on V . But we saw in the proof of (c) thatλ(H) = 1

dimVλtrVλ H, and h is a subalgebra of g′, and cyclicicty assures that

tr[X,Y ] = 0. Hence λ(H) = 0 for all H ∈ h. But then h acts as 0 on Vλ, and soannihilates and simple. Thus h ⊆ n. **

5. Let g be a finite-dimensional Lie algebra over K, char(K) = 0. Prove that the largest nilpotentideal of g is equal to the set of elements of r = rad g which act nilpotently in the adjoint actionon g, or equivalently on r. In particular, it is equal to the largest nilpotent ideal of r.

**Let n be the largest nilpotent ideal of g. Any nilpotent ideal of g acts nilpotentlyon g, hence n is the largest ideal which acts nilpotently in the adjoint action. Letβ be the Killing form; then n ⊆ radβ ⊆ rad g. (The first inequality is a corollary ofEngel’s theorem. The second is a corollary of Cartan’s criterion.)

It suffices to show that the largest nilpotent ideal of r = rad g is an ideal of g. Thisfollows from Levi’s theorem and general nonsense: g = sn r, and so g acts on r byinfinitesimal automorphisms. But the largest nilpotent ideal of r is characteristicin r, so fixed by any automorphism. **

6. Prove that the Lie algebra sl(2,K) of 2× 2 matrices with trace zero is simple, over a field Kof any characteristic 6= 2. In characteristic 2, show that it is nilpotent.

**A basis of g = sl(2,K) is

E =Ç

0 10 0

å, F =

Ç0 01 0

å, and H =

Ç1 00 −1

å(PS4.12)

The brackets are [E,F ] = H, [H,E] = 2E, and [H,F ] = −2F . When K is characteristic-2, then H is central, [g, g] = KH, and [g,KH] = 0, so g is nilpotent; indeed, g is aHeisenberg algebra.

When charK 6= 2, let a be an ideal of g and X = eE + fF + hH ∈ a for e, f, h ∈ K.Then [E, [E,X]] = [E, fH − 2hE] = −2fE and [F, [F,X]] = [F,−eH + 2hF ] = −2eF . SoE ∈ a unless f = 0 and F ∈ a unless e = 0. If e = f = 0, then H ∈ a unless h = 0. Soif X 6= 0, then a contains at least one of E,F,H, but any one of these generatesthe rest by brackets. Hence a = g is simple. **

7. In this exercise, well deduce from the standard functorial properties of Ext groups and theirassociated long exact sequences that Ext1(N,M) bijectively classifies extensions 0 → M →V → N → 0 up to isomorphism, for modules over any associative ring with unity.

(a) Let F be a free module with a surjective homomorphism onto N , so we have an exactsequence 0→ K → F → N → 0. Use the long exact sequence to produce an isomorphismof Ext1(N,M) with the cokernel of Hom(F,M)→ Hom(K,M).

**The long exact sequence in Ext(−,M) coming from the short-exact 0 →

165

K → F → N → 0 begins

0→ Ext0(N,M)→ Ext0(F,M)→ Ext0(K,M)→ Ext1(N,M)→ Ext1(F,M)→ . . .(PS4.13)

But Ext0 = Hom by construction, and Ext1(F,M) = 0 since F is free, soin particular Hom(F,M) → Hom(K,M) → Ext1(N,M) → 0 is exact, and soExt1(N,M) = coker{Hom(F,M)→ Hom(K,M)}. **

(b) Given φ ∈ Hom(K,M), construct V as the quotient of F ⊕M by the graph of −φ (notethat this graph is a submodule of K ⊕M ⊆ F ⊕M).

**If φ ∈ Hom(K,M), then (a,−φ) : K → F ⊕M traces out a submodule Φ ofF ⊕M , where a is the map from K → F in the exact sequence in (a). LetV = (F⊕M)/Φ, and b : F⊕M → V the cokernel of (a,−φ). Then b◦(0, id) : M → Vis injective; if m 7→ 0, then b : (0,m) 7→ 0, so m = φ(0) = 0. On the other hand,we construct the cokernel V/M of this injection by taking pairs f⊕m ∈ F ⊕M ,modding out by k⊕−φ(k) = 0 for k ∈ K, and then modding out by the secondcoordinate. We could do the modding out in the other order, in which casewe see that V/M ∼= F/K ∼= N canonically. In more categorical language, V/Mis the cokernel of

Ka //φ

$$IIIIIII F⊕ ⊕M

id // M

(PS4.14)

in which we can perform the M quotient first if we choose, yielding F/K ∼= N .Thus 0→M → V → N → 0 is exact. **

(c) Use the functoriality of Ext and the long exact sequences to show that the characteristicclass in Ext1(N,M) of the extension constructed in (b) is the element represented by thechosen φ, and conversely, that if φ represents the characteristic class of a given extension,then the extension constructed in (b) is isomorphic to the given one.

**We recall that any extension 0 → M → V → N → 0 picks out an elementof Ext1(N,M), by writing out the long-exact sequence and tracing the imageof id ∈ Hom(M,M) in Ext1(N,M). Let V depend on φ as in part (b). Byconstruction, the following squares commute, with exact rows:

0 // K

φ

��

// F

��

// N // 0

0 // M // V // N // 0

(PS4.15)

We now pass this to the long-exact sequences in Ext:

// Hom(F,M) // Hom(K,M) // Ext1(N,M) //

// Hom(V,M) //

OO

Hom(M,M) //

φ

OO

Ext1(N,M) //

(PS4.16)

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But chasing id ∈ Hom(M,M) in the right square gives

φ � // equivalence class of φ

id_

OO

� // characteristic class of V

(PS4.17)

Hence the characteristic class of the extension constructed in (b) is the imageof φ in Ext1(N,M).

Conversely, let 0 → M → V → N → 0 be any extension. Since F is free, wecan find a map F → V so that the right square commutes:

0 // K // F //

�� N // 0

0 // M // V // N // 0

(PS4.18)

But K is the kernel of the map F → N , and so maps into the kernel of V → Nunder the map F → V . Thus the map F → V restricts to a map φ : K → M ,i.e. each square in the following diagram commutes:

0 // K //

φ

��

F //

��

N // 0

0 // M // V // N // 0

(PS4.19)

Thus the following sequence is exact:

0→ K → F ⊕M → N ⊕ V → N → 0 (PS4.20)

Labeling the maps a : K → F and c : V → N , then the map K → F ⊕M is(a,−φ); the map N ⊕ V → N is (id,−c). Then the kernel of N ⊕ V → N is thegraph of c : V → N in V ⊕ N , and hence is isomorphic to V (any graph isisomorphic to its domain). Thus 0 → K → F ⊕M → V → 0 is exact, and V isone of the extensions constructed in part (b).

It suffices to show that the extensions constructed in part (b) are classifiedby their characteristic class. Using part (a), we want to show that if ψ ∈Hom(F,M) and φ ∈ Hom(K,M), then the extensions constructed from φ andφ+ψ are isomorphic. But in (PS4.18) we could have changed the map F → Vby any map F →M ; this changes φ by the same amount. This completes theproof. **

8. Calculate Exti(K,K) for all i for the trivial representation K of sl(2,K), where char(K) = 0.Conclude that the theorem that Exti(M,N) = 0 for i = 1, 2 and all finite-dimensionalrepresentations M,N of a semi-simple Lie algebra g does not extend to i > 2.

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**Let U = U(sl(2)). We recall that

· · · → U ⊗∧3sl(2)→ U ⊗∧2sl(2)→ U ⊗ sl(2)→ U (PS4.21)

is a free resolution of the trivial module sl(2) y K. We take HomU (−,K) and cancelthe U⊗ with the HomU ; hence, we want to calculate the homology of the complex

K→ HomK(sl(2),K)→ HomK(∧2sl(2),K)→ HomK(

∧3sl(2),K)→ HomK(∧4sl(2),K)→ . . .

(PS4.22)But sl(2) is three-dimensional over K, so

∧isl(2) = 0 for i ≥ 4. Thus, the onlynon-zero homologies can live in dimensions 0, . . . , 3, and we are interested in thesequence (where A∗ = Hom(A,K) is the normal dual space)

0→ K δ1→ sl(2)∗ δ2→Ä∧2sl(2)

ä∗ δ3→Ä∧3sl(2)

ä∗ → 0 (PS4.23)

At dimension 0, Ext0(K,K) = HomU (K,K) = K, and at dimension 1 we have alreadycomputed, through other means, that Ext1(K,K) = 0. For these to be true, δ1 mustbe the 0 map, and δ2 must be an injection. But dim

(∧2sl(2))∗ = dim sl(2)∗ = 3, so δ2

is onto, hence δ3 must be the zero map and Ext2(K,K) = 0. Lastly, dim(∧2sl(2)

)∗ =1, so Ext3(K,K) = K/0 = K.

We can compute these directly. Since x · a = 0 for any x ∈ sl(2) and a ∈ K (where ·is the action sl(2) y K), we have an explicit description of the boundary maps:

δk(g)(x1 ∧ · · · ∧ xk) =∑

(−1)i+jg([xi, xj ] ∧ x1 ∧ · · · xi · · · xj · · · ∧ xk) (PS4.24)

Sure enough, δ1 = 0. δ2 is dual to the bracket [, ] :∧2sl(2) → sl(2), which is an

isomorphism since sl(2) is simple. δ3 is dual to the map

e ∧ h ∧ f 7→ −[e, h] ∧ f + [e, f ] ∧ h− [h, f ] ∧ e (PS4.25)= 2e ∧ f + h ∧ h+ 2f ∧ e (PS4.26)= 0 (PS4.27)

as predicted in the previous paragraph. **

9. Let g be a finite-dimensional Lie algebra. Show that Ext1(K,K) can be canonically identifiedwith the dual space of g/g′, and therefore also with the set of 1-dimensional g-modules, upto isomorphism.

**Again we use the fact that (in characteristic 0) for any g,

· · · → U(g)⊗∧3g→ U(g)⊗∧2g→ U(g)⊗ g→ U(g) (PS4.28)

is a free-resolution of g y K, and hence Exti(K,K) is the homology of

K→ g∗ →Ä∧2g

ä∗ → Ä∧3gä∗ → . . . (PS4.29)

168

where we write A∗ for HomK(A,K). Since g y K trivially, the boundary mapsδk :Ä∧k−1g

ä∗ → Ä∧kgä∗ are dual to the maps

dk : x1 ∧ · · · ∧ xk 7→∑i<j

(−1)i+j [xi, xj ] ∧ x1 ∧ · · · xi · · · xj · · · ∧ xk (PS4.30)

In any case, we want to compute Ext1(K,K) = ker δ2, since δ1 : K → g∗ is the zeromap. But d2 is just the (negative of the) bracket x1 ∧ x2 7→ −[x1, x2] :

∧2g→ g. I.e.(δ2f)(x∧y) = −f([x, y]) for x, y ∈ g. So f ∈ ker δ2 = Ext1(K,K) if and only if f |g′ = 0, i.e.if and only if f factors through g→ g/g′. Thus Ext1(K,K) = HomK(g/g′,K) = (g/g′)∗

is the space of (kernels of) one-dimensional modules of g/g′, since any such moduleis given by a map g/g′ → gl(K) = K. Of course, any one-dimensional module of g

is an abelian module, so Ext1(K,K) classifies one-dimensional modules of g up toisomorphism. **

10. Let g be a finite-dimensional Lie algebra. Show that Ext1(K, g) can be canonically identifiedwith the quotient Der(g)/ Inn(g), where Der(g) is the space of derivations of g, and Inn(g)is the subspace of inner derivations, that is, those of the form d(x) = [y, x] for some y ∈ g.Show that this also classifies Lie algebra extensions g containing g as an ideal of codimension1.

**Exti(K, g) is computed by

gδ1→ Hom(g, g) δ2→ Hom(

∧2g, g)→ . . . (PS4.31)

where (δ1h)(x) = x · h and (δ2f)(x ∧ y) = x · f(y) − y · f(x) − f([x, y]), for x, y ∈ g andh ∈ g, f ∈ Hom(g, g). The action · is the bracket [, ], so

(δ1h)(x) = [x, h], and (PS4.32)

(δ2f)(x, y) = [x, f(y)] + [f(x), y]− f([x, y]). (PS4.33)

We have Ext1(K, g) = ker δ2/ im δ1. But ker δ2 = Der(g) is the space of derivationsof g, i.e. mapst f : g → g such that the right-hand side of (PS4.33) is 0. Thenδ1 : g→ Der g by h 7→ − adh; by definition, im δ1 = Inn(g).

We now show explicitly how Der(g)/ Inn(g) classifies Lie algebra extensions g ofg containing g as a one-codimensional ideal. Let 0 → g → g → K → 0 be onesuch extension, and let x ∈ g r g. Then since g is an ideal of g, adx acts on g

as a derivation; since g is co-dimension 1, this derivation and the structure of g

completely determines the structure of g. Let y be any other representative ofthe image of x in K; then x − y ∈ g and so the derivation adx − ad y is in Inn g;hence to each extension 0 → g → g → K → 0 such that g is an ideal in g we canassociate an element of Der(g)/ Inn(g).

Conversely, let ξ ∈ Der(g)/ Inn(g), and ξ ∈ Der(g) a representative derivation. Thenwe can construct an extension 0 → g → g → K → 0 by letting g = g ⊕ Kx as

169

vector spaces for x a formal symbol, and [x, y] = ξ(y) when y ∈ g. Then g isa subalgebra and moreover an ideal, since ξ(y) ∈ g if y ∈ g. Let ξ be anotherrepresentative in Der(g) of ξ. Then ξ − ξ ∈ Inn(g), and let’s say y ∈ Inn(g) so thatξ − ξ = ad y. Let g = g ⊕ Kx be the extension built from ξ. Consider the mapg → g that fixes g and sends x 7→ x + y. This is a bijection of K-vector spacesand trivially an isomorphism on the subalgebra g. But in fact if z ∈ g, then[x, z]g = ξ(z) = ξ(z) + (ad y)(z) = ξ(z) + [y, z] = [x + y, z]g, so the map g → g is ahomomorphism of Lie algebras.

Thus g ∼= g, and Der(g)/ Inn(g) classifies Lie algebra extensions 0 → g → g → K → 0such that g is an ideal of g. **

11. Let g be a finite-dimensional Lie algebra. Show that there is a canonical isomorphismExt1(g,K) ∼= Ext2(K,K) ⊕ S2 ((g/g′)∗) where S2 denotes the second symmetric power. Thefirst term classifies those g-module extentions 0→ K→ g→ g→ 0 that are (one-dimensional,central) Lie algebra extensions.

Addendum: This problem turned out to be harder than I thought, and I’m not even surethat its true.

Lets assume the ground field has char(K) 6= 2, so we can distinguish between symmetric andantisymmetric forms.

The weaker result that there is a canonical injection Ext2(K,K)⊕ S2((g/g′)∗) ↪→ Ext1(g,K)can be proven by representing a 1-cocyle as a bilinear form on g and considering the caseswhere the form is antisymmetric or symmetric.

For the stronger result, note that the identity ([x, z], z) = 0 holds for the symmetriza-tion of the form representing a 1-cocycle. Then ([x, y], z) + (y, [x, z]]) = ([x, y + z], y +z)?([x, y], y)?([x, z], z) = 0, so the symmetrized form is invariant. Among the invariant sym-metric forms are those whose radical contains g′. These represent 1-cocycles. But somefurther argument is needed to show that no other invariant form can arise by symmetrizinga 1-cocyle.

12. Let g be a finite-dimensional Lie algebra over K, char(K) = 0. The Malcev-Harish-Chandratheorem says that all Levi subalgebras s ⊆ g are conjugate under the action of the groupexp ad n, where n is the largest nilpotent ideal of g (note that n acts nilpotently on g, so thepower series expression for exp adX reduces to a finite sum when X ∈ n).

(a) Show that the reduction we used to prove Levi’s theorem by induction in the case wherethe radical r = rad g is not a minimal ideal also works for the Malcev-Harish-Chandratheorem. More precisely, show that if r is nilpotent, the reduction can be done usingany nonzero ideal m properly contained in r. If r is not nilpotent, use Problem 4 to showthat [g, r] = r, then make the reduction by taking m to contain [g, r].

(b) In general, given a semidirect product g = hnm, where m is an abelian ideal, show thatExt1

U(h)(K,m) classifies subalgebras complementary to m, up to conjugacy by the action

170

of exp ad m. Then use the vanishing of Ext1(M,N) for finite-dimensional modules overa semi-simple Lie algebra to complete the proof of the Malcev-Harish-Chandra theorem.


1. (a) Show that SL(2,R) is topologically the product of a circle and two copies of R, hence itis not simply connected.

**The subgroup SO(2,R) ⊆ SL(2,R) of matrices of the formÇ

cos θ sin θ− sin θ cos θ

åacts freely on SL(2,R) by right-multiplication, say. Then the orbit of any

matrix is a circle, since SO(2,R) ∼= S1. In particular, the orbit ofÇa bc d

åare all matrices of the form

Ça cos θ − b sin θ a sin θ + b cos θc cos θ − d sin θ c sin θ + d cos θ

åfor a, b, c, d fixed

and θ ∈ S1. For any given c, d ∈ R, there are exactly two θ ∈ S1 such thatc cos θ−d sin θ = 0, namely the θ with tan θ = c/d. Then c sin θ+d cos θ = ±

√c2 + d2,

and there is exactly one θ ∈ S1 with c cos θ − d sin θ = 0 and c sin θ + d cos θ >0. Thus each SO(2,R)-orbit in SL(2,R) intersects the subgroup U of upper-triangular matrices with positive diagonal entries exactly once, and at leasttopologically SL(2,R)/SO(2,R) ∼= U . But U ∼= R2, since we we can parameterize

it as U = {Çet b0 e−t

å: t, b ∈ R}. **

(b) Let S be the simply connected cover of SL(2,R). Show that its finite-dimensional com-plex representations, i.e., real Lie group homomorphisms S → GL(n,C), are determinedby corresponding complex representations of the Lie algebra Lie(S)C = sl(2,C), andhence factor through SL(2,R). Thus S is a simply connected real Lie group with nofaithful finite-dimensional representation.

**By the usual Lie group magic, any map S → GL(n,C) is determined by amap of the associated real lie algebras sl(2,R)→ gl(n,C). Although gl(n,C) is(2n2)-dimensional as a real Lie algebra, we can think of any map sl(2,R) →gl(n,C) as a 3 × n2 complex-valued matrix, and so it determines a C-linearmap sl(2,C) → gl(n,C) which agrees with the map sl(2,R) → gl(n,C) on thereal subalgebra sl(2,R) ⊆ sl(2,C). But the structure constants of sl(2,R) and

sl(2,C) are the same; each has a a basis e =Ç

0 10 0

å, f =

Ç0 01 0

å, and

h =Ç

1 00 −1

å, where [e, f ] = h, [f, h] = 2f , and [h, e] = 2e. Thus the map

sl(2,C) → gl(n,C) is a complex Lie algebra homomorphism if and only if themap sl(2,R)→ gl(n,C) is a real Lie algebra homomorphism.

171

Thus any real Lie group homomorphism S → GL(n,C) determines a com-plex Lie algebra homormorphism sl(2,C) → gl(2,C). Since SL(2,C) is simply-connected, the map sl(2,C)→ gl(2,C) integrates to a map SL(2,C)→ GL(2,C).Restricting to the real subgroup SL(2,R) ⊆ SL(2,C), we get a representationof SL(2,R) with the same infinitesimal representation sl(2,R) → gl(n,C) thatour original function S → GL(n,C) had. So the original map factors throughS �6∼SL(2,C), and must not have been faithful. **

2. (a) Let U be the group of 3 × 3 upper-unitriangular complex matrices. Let Γ ⊆ U be thecyclic subgroup of matrices 1 0 m

0 1 00 0 1

,where m ∈ Z. Show that G = U/Γ is a (non-simply-connected) complex Lie group thathas no faithful finite-dimensional representation.

**We remark first that Γ is central and hence normal:Ö1 m

11

èÖ1 a b

1 c1

è=

Ö1 a b

1 c1

èÖ1 m

11

è=

Ö1 a b+m

1 c1

è(PS5.1)

Thus

U/Γ =

Ö

1 a [b]1 c

1

ès.t. a, c ∈ C, [b] ∈ C/Z

(PS5.2)

Now let φ : U/Γ → GL(n,C) be a Lie group homomorphism, and dφ : u →gl(n,C) its infinitesimal Lie algebra homomorphism. We remark that u is thecomplexified Heisenberg algebra: a (C-)basis of u are the derivatives ∂

∂a ,∂∂b ,

and ∂∂c , whence ∂

∂b is central and [ ∂∂a ,∂∂c ] = ∂

∂b .

In U , Ö1 b

11

è= exp

Åb∂

∂b

ã(PS5.3)

and so

φ

Ö1 b

11

è= exp

Åb dφ

Å∂

∂b

ãã(PS5.4)

But dφB is nilpotent in gl(n,C), so (the matrix entries of) exp(b dφB) is apolynomial f(b) in b. If b = m ∈ Z, then the left-hand side of (PS5.4) is theidentity, since φ : U → GL(n,C) factors through U/Γ, and so f(b) takes the

172

same value infinitely often, hence f(b) must be constantly the identity inGL(n,C). Hence dφ cannot have full rank, and φ is not faithful. **

(b) Adapt the solution to Set 4, Problem 2(b) to construct a faithful, irreducible infinite-dimensional linear representation V of G.

**A faithful representation of u in which ∂∂b is not nilpotent is u y C[x], where

∂∂b 7→ id, ∂

∂a 7→ddx , and ∂

∂c 7→ x×. This exponentiates to an action U y C[x] inwhich Ö

1 b1

1

è7→ eb× (PS5.5)

This does not factor through U/Γ — em 6= 1 for m ∈ Z — but a slight mod-ification does. To wit, we let u y C[x] by ∂

∂a 7→ 2πi ddx , and ∂∂c 7→ x×. The

∂∂b = [ ∂∂a ,

∂∂c ] = 2πi×, and we integrate to an action U y C∞(C) in whichÖ

1 b1

1

è7→ e2πib× (PS5.6)

In particular, the kernel of this action includes Γ, since e2πim = 1 if m ∈ Z.But Ö

1 a1

1

è7→ shift by a (PS5.7)

and Ö1

1 c1

è7→ e2πicx× (PS5.8)

are never trivial unless a and c are each 0, and most generallyÖ1 a b

1 c1

è=

Ö1

1 c1

èÖ1 b

11

èÖ1 a

11

è: f(x) 7→ e2πicx+2πibf(x− a)

(PS5.9)also doesn’t fix all functions unless a = c = 0 and b ∈ Z. So the kernel ofU y C[x] is exactly Γ, and we have a faithful representation of U/Γ. **

3. Following the outline below, prove that if h ⊆ gl(n,C) is a real Lie subalgebra with theproperty that every X ∈ h is diagonalizable and has purely imaginary eigenvalues, then thecorresponding connected Lie subgroup H ⊆ GL(n,C) has compact closure (this completesthe solution to Set 1, Problem 7).

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(a) Show that adX is diagonalizable with imaginary eigenvalues for every X ∈ h.

**Let {ei} be a basis in which X is diagonal: Xei = Xiei (no sum) for Xi

pure imaginary. Then we get a basis {eji} of gln(C), and adX : eji 7→ [X, eji ] =(Xi −Xj)e

ji is diagonal with pure-imaginary eigenvalues. **

(b) Show that the Killing form of h is negative semidefinite and its radical is the center ofh. Deduce that h is reductive and the Killing form of its semi-simple part is negativedefinite. Hence the Lie subgroup corresponding to the semi-simple part is compact.

**Let β be the killing form on h, i.e. β(X,Y ) = trh(adX adY ). Then adX isdiagonalizable with pure-imaginary eigenvalues, so (adX)2 is diagonalizablewith non-positive eigenvalues, so β(x, x) is a sum of non-positive numbers andhence non-positive. If adX 6= 0, then β(X,X) < 0, and so radβ can consistonly of X ∈ h such that adX = 0 (of course, radβ contains all such X), andso radβ = {X ∈ h s.t. adX = 0} = Z(h) is the center of h. Write z = Z(h) andg = h/z, Z for the subgroup corresponding to z (necessarily in the center ofH), and G = H/Z. Then Lie(G) = g.

In any case, the adjoint action h y h factors through g, which is the point,and so h is a g-module, and indeed an extension of the form

0→ z→ h→ g→ 0 (PS5.10)

where g y z trivially. But the adjoint action h y g also factors through g,and the trace form of this action is exactly the killing form of h. So g issemisimple, and Ext1(z, g) = 0. Thus (PS5.10) is the trivial extension, and h

is reductive.

In any case, as a topological space H is a quotient of G×Z. And ad : g ↪→ gl(g),but in fact, letting βg be the Killing form on g, we see that ad : g ↪→ so(g, β).In particular, Ad : G → SO(g, β), and since β is negative-definite, SO(g, β) iscompact.

Let GR be the simply-connected cover of G; it depends only on g. Whenwe complexify, we get gC = g⊗R C, with simply-connected cover GC, and thefollowing diagram commutes (the top square is made of Lie algebras, the

174

bottom are Lie groups):

g

exp

��O�O�O�O�O�O�O�O�O�O

� o

��?????????ad // soR(g, β)

�O�O�O�O

��O�O�O�O

� o

��????????

gC


// soC(gC)

exp


GR//

� o

��????????SOR(g, β)

� o

��????????

GC// SOC(gC)

(PS5.11)

The maps out of GC,R exist because they are simply-connected, and the mapGR → GC is an embedding because GC is simply connected and g ↪→ gC. In anycase, gC is semisimple over C, and we know the classification of groups with agiven semisimple Lie algebra. In particular, the image of GC in SOC(gC) is afinite quotient of GC, and so, since the bottom square commutes, the imageof GR in SOR(g, β) is also a finite quotient of GR. But this image is compact,so GR is compact, and so any other quotient of it, and in particular the Gfrom the previous paragraph, is also compact. **

(c) Show that the Lie subgroup corresponding to the center of h is a dense subgroup of acompact torus. Deduce that the closure of H is compact.

**z is an abelian Lie algebra, all of whose elements are diagonalizable withpure-imaginary eigenvalues. But if a collection of diagonalizable matricescommute, then there is a basis that simultaneously diagonalizes all of them;in this bases, z is an algebra of diagonal matrices all of whose entries arepure-imaginary numbers, and z ⊆ (iR)n ⊆ gl(n,C). (Recall that z ⊆ h ⊆ gl(n,C)to begin with.) Then

Z = exp(z) ⊆

eiti

. . .eitn

s.t. t1, . . . , tn ∈ R

= Tn ⊆ GL(n,C) (PS5.12)

which is compact. Of course, Z might not be dense in Tn, but the closure ofZ is a closed subgroup of Tn, and these are all tori.

Recall that h = z× g, and G is compact. Then H = ZG ⊆ GL(n,C), and Z hascompact closure, so H does. **

175

(d) Show that H is compact — that is, closed — if and only if it further holds that thecenter of h is spanned by matrices whose eigenvalues are rational multiples of i.

**H is compact if and only if Z is: Z is a closed subgroup of H, because beingcentral is a closed condition. But Z ⊆ T is compact if and only if it is closed.

Consider the map e : z→ Tn, the restriction of e : (it1, . . . , itn) 7→ (e2πit1 , . . . , e2πitn) :(iR)n → Tn. The kernel ker e is necessarily a discrete subgroup of z. Then ker eis full-rank as a lattice if and only if Z = z/ ker e is a compact torus. Butker e = iZn ∩ z, so if ker e is full-rank, then z is spanned by matrices all ofwhose eigenvalues are integer multiples of i. Conversely, let z be spannedby matrices all of whose eigenvalues are rational multiples of i; then up tomultiplication by an integer, z is spanned by matrices all of whose eigenvaluesare integer multiples of i, and ker e is full-rank. **

4. Let Vn = Sn(C2) be the (n+ 1)-dimensional irreducible representation of sl(2,C).

(a) Show that for m ≤ n, Vm ⊗ Vn ∼= Vn−m ⊕ Vn−m+2 ⊕ · · · ⊕ Vn+m, and deduce that thedecomposition into irreducibles is unique.

**I would like to remark first that I’m not sure why the uniqueness of the de-composition into irreducibles follows from the identity Vm⊗Vn =

⊕md=0 Vm+n−2d

when m < n. Indeed, let C be a semisimple K-linear category, and A =⊕Vi =⊕

Wj be two decompositions of an object into simples. Then each Vi mustmap into

⊕Wj, and Hom(Vi,Wj) = 0 unless Vi ∼= Wj. So the list of simples

in the decompositions are the same up to multiplicity. So we need onlythink about repeated sums of the same simple; can we have (⊕n)V ∼= (⊕m)Vfor a simple V and n 6= m? No. Indeed, ⊕ pulls out of Hom, and soHom(V, (⊕n)V ) = (⊕n) Hom(V, V ), which is an n-dimensional K-vector space.Since dimK is well-defined in K-linear categories, we must have n = m if(⊕n)V ∼= (⊕m)V . Thus decomposition into simples in a semisimple categoryis unique. And we have shown that sl(2,C)-representations form a semisimplecategory: every irreducible is simple.

Let x, y be a basis for C2, on which sl(2,C) = 〈e, f, h〉 acts as

e =®x 7→ 0y 7→ x

´, x =

®x 7→ yy 7→ 0

´, and h =

®x 7→ xy 7→ −y

´(PS5.13)

We recall that if g ∈ g and V,W are two g-modules, then g acts on V ⊗W byg ⊗ 1 + 1 ⊗ g. A basis for the symmetric tensor product Sn(C2) = Vn are thehomogeneous polynomials {xn−jyj , 0 ≤ j ≤ n}. Then e : xn−jyj 7→ jxn−j+1yj−1 =x ∂∂yx

n−jyj, f acts as y ∂∂x , and h : xn−jyj 7→ (n− 2j)xn−jyj.

Let Vm be the same as Vn with the letters (x, y, n, j) replaced by (a, b,m, i).Then Vm ⊗ Vn is spanned by the monomials aibm−ixjyn−j for 0 ≤ i ≤ m and

176

0 ≤ j ≤ n. Then e = a ∂∂b +x ∂

∂y , f = b ∂∂a +y ∂∂x , and h acts diagonally, multiplying

aibm−ixjyn−j by m+ n− 2(i+ j).

We remark that e takes any polynomial p homogeneous in “ax-degree” dega p+degx p = i+j to a polynomial homogeneous in ax-degree i+j+1, and multipliesthe sum of the coefficients thereof by (m + n) − (i + j). The picture of theaction of e is this (f has the opposite picture):

i //

0 1 2 . . . m− 1 m

j

��

0 • m //

n��

•m−1//

n��

•m−2//

n��

. . . 2 // • 1 //

n��

•n

��1 • m //

n−1��

•m−1//

n−1��

•m−2//

n−1��

. . . 2 // • 1 //

n−1��

•n−1��

2 • m //

n−2��

•m−1//

n−2��

•m−2//

n−2��

. . . 2 // • 1 //

n−2��

•n−2��

3 • m //

n−3��

•m−1//

n−3��

•m−2//

n−3��

. . . 2 // • 1 //

n−3��

•n−3��

4 • m //

n−4��•m−1//

n−4��•m−2//

n−4��

. . . 2 // • 1 //

n−4��•n−4��

......

2��

...2��

...2��

. . ....

2��

...2��

n− 1 • m //

1��

•m−1//

1��

•m−2//

1��

. . . 2 // • 1 //

1��

•1��

n • m // •m−1// •m−2// . . . 2 // • 1 // •

(PS5.14)

In the above picture, the (i, j)th dot corresponds to the monomial aibm−ixjyn−j.The action of e on a monomial is the sum over arrows leaving that monomialof the target of the arrow; of course, an arrow labeled by k is really k arrows.The “ax-degree” dega + degx of a monomial is its inverse-diagonal; i.e. we’regrading the module in the diagonal direction.

Let p be a polynomial in ker e; then it is the sum of polynomials each ho-mogeneous in i + j, and each of these must be in the kernel. So let p be apolynomial all of whose monomials have the same value of dega + degx; thenp is a weighting on an inverse diagonal. It’s easy enough to see that ifi + j < n − m, then p cannot be in the kernel of e. To wit, let k < n andp =

∑i+j=k pia

ibm−ixjyn−j. Let i1 be the smallest i such that pi 6= 0. Thene(p) = (n − k + i1)pi1a

ibm−ixk−i1+1yn−k+i1−1 + . . . , where all other terms have alarger degree in a. On the other hand, for each n ≤ k ≤ n+m, there is exactlyone polynomial (up to a scalar) of ax-degree i+ j = k in ker p. In particular:

n+m−k∑l=0

(−1)lÇn+m− k

l

åak−n+lbn+m−k−lxn−lyl ∈ ker e (PS5.15)

177

using the fact that (r − s)(rs

)= (s+ 1)

( rs+1

)= r!

s!(r−s−1)! for any integers r, s.

To save time, write d = n+m− k. Then:

ker e 3n+m−k∑l=0

(−1)lÇn+m− k

l

åak−n+lbn+m−k−lxn−lyl (PS5.16)

= am−dxn−dd∑l=0

(−1)lÇd

l

åalbd−lxd−lyl (PS5.17)

= am−dxn−d (ay − bx)d (PS5.18)

We remark that (ay−bx) is in the kernel of both e and f , which are derivations,so e[p · (ay − bx)] = e[p] · (ay − bx), and similarly for f . In any case, for eachd ∈ 0, . . . ,m, we get a vector in ker e on which h acts as −(m + n − 2d). Sureenough, too, it takes n+m− 2d iterations of f to annihilate this vector, andit’s clear that the vector generates a submodule of dimension m+n−2d. But(m+1)(n+1) =

∑nd=0m+n−2d+1, and the sum of these submodules generates

Vm⊗Vn hence is the whole module, so the submodules are completely linearlyindependent, and we have exhibited the identity

Vm ⊗ Vn =m⊕d=0

Vm+n−2d (PS5.19)

**

(b) Show that in any decomposition of V ⊗n1 into irreducibles, the multiplicity of Vn is equalto 1, the multiplicity of Vn−2k is equal to

(nk

)−( nk−1

)for k = 1, . . . , bn/2c, and all other

irreducibles Vm have multiplicity zero.

**We proceed by induction, using the fact that Vm ⊗ V1 = Vm+1 ⊗ Vm−1, andthat ⊗ distributes over ⊕. Then, understanding

( n−1

)=( n−2

)= 0 for n ≥ 0, and

178

writing mV for the direct sum of m copies of V , we have

V ⊗n1 =bn−1

2 c⊕k=0

ñÇn− 1k

å−Çn− 1k − 1

åôVn−1−2k ⊗ V1 (PS5.20)

=bn−1

2 c⊕k=0

ñÇn− 1k

å−Çn− 1k − 1

åô(Vn−2k ⊕ Vn−2k−2) (PS5.21)

=bn−1

2 c⊕k=0

ñÇn− 1k

å−Çn− 1k − 1

åôVn−2k ⊕

bn−12 c⊕

k=0

ñÇn− 1k

å−Çn− 1k − 1

åôVn−2(k+1)

(PS5.22)

=bn−1

2 c⊕k=0

ñÇn− 1k

å−Çn− 1k − 1

åôVn−2k ⊕

bn+12 c⊕

k=0

ñÇn− 1k − 1

å−Çn− 1k − 2

åôVn−2k

(PS5.23)

=

[Çn− 1ön−1

2

ùå− Ç n− 1ön−1

2

ù− 1

å]Vn−2bn+1

2 c ⊕bn−1

2 c⊕k=0

ñÇn

k

å−Ç

n

k − 1

åôVn−2k

(PS5.24)

since(nk

)+( nk−1

)=(n+1k

)and we telescope the sum. When n is odd,

ön−1

2

ù=⌊n

2

⌋,

and n−2ön+1

2

ù= n−2n+1

2 = −1, and V−1 is the zero-dimensional representation.When n is even,

⌊n2

⌋= 1 +

ön−1

2

ù, and

( n−1

bn−12 c)−( n−1

bn−12 c−1

)=(n−1n2−1

)−(n−1n2−2

)=(n−1

n2

)−(n−1n2−2

)=(n−1

n2

)+(n−1

n2

)−(n−1

n2

)−(n−1n2−2

)=(nn2

)−( nn2−1

). In either case, the

right-hand side of (PS5.24) is equal to⊕bn/2c

k=0

î(nk

)−( nk−1

)óVn−2k. **

5. Let A be a symmetric Cartan matrix, i.e., A is symmetric with diagonal entries 2 and off-diagonal entries 0 or −1. Let Γ be a subgroup of the automorphism group of the Dynkindiagram D of A, such that every edge of D has its endpoints in distinct Γ orbits. Define thefolding D′ of D to be the diagram with a node for every Γ orbit I of nodes in D, with edgeweight k from I to J if each node of I is adjacent in D to k nodes of J . Denote by A′ theCartan matrix with diagram D′.

(a) Show that A′ is symmetrizable and that every symmetrizable generalized Cartan matrix(not assumed to be of finite type) can be obtained by folding from a symmetric one.

**A matrix aij is symmetrizable if and only if it satisfies two conditions:(i) that aij = 0 exactly when aji = 0 for any pair i, j, and (ii) that for anysequence i1, . . . , ik, ai1,i2 . . . aik−1,ikaik,i1 = ai1,ikaik,ik−1

. . . ai2,i1. For a folding A′ ofa symmetric Cartan A, condition (i) is immediate: if there is at least oneedge from I to J , then there is at least one from J to I. As for condition (ii),A′IJA

′JK counts the number of paths in D from a particular node in I to any

node in J to any node in K, and thus is 1/|I| times the number of paths from

179

a node in I to a node in K, and so (−1)k|I1|A′I1,I2 . . . A′Ik,I1

counts the numberof paths from some node in I1 to a node in I2, etc. returning to I1. But anysuch path is reversible, and so (−1)k|I1|A′I1,Ik . . . A

′I2,I1

counts the same thing.

Let D′ be a symmetrizable Dynkin diagram, which we assume to be con-nected (disconnected diagrams can be unfolded separately). We count di-rected paths: an edge labeled by k has k paths going along the arrow and 1path going against the arrow. Then for any sequence of vertices I1, I2, . . . , Ik,the number of paths going I1, I2, . . . , Ik, I1 must equal the number of pathsI1, Ik, . . . , I2, I1. Thus, let I, J be two vertices of D′ connected by an edge ofweight k ≥ 2, so that there are k edges from I to J . Then for any other se-quence of edges starting at I and ending at J , there must be k times as manypaths going along the sequence as going against. Thus we build a partialunfolding D of D′ by expanding J to k-many nodes, each connected singly toI, and for any other way to get from I to J , expanding any directed edgesso that at the end there are k branches. In D, we consider two edges equiv-alent if they fold to the same edge in D′; by construction, two equivalentedges have the same weight, and there are fewer equivalence classes in Dwith weight at least two than there are in D′. Thus, we can partially unfoldD again — by counting paths, it’s clear that D is still symmetrizable — byunfolding along every edge in an equivalence class. Rinse, repeat, and we’redone by induction on the number of equivalence classes of edges of weight atleast two. **

(b) Show that every folding of a finite type symmetric Cartan matrix is of finite type.

**The Dynkin diagram of a finite-type symmetric Cartan matrix consistsof the disjoint union of diagrams of types A, D, and E. No automorphismcan identify non-isomorphic components, and if an autormophism identifiestwo nodes in different components, then it identifies the whole components.But since there were originally no paths from one component to another,a folding of a disconnected Dynkin diagram depends only on the inducesautomorphisms of each component in the preimage of a connected componentof the folded diagram. Thus, to understand foldings of finite-type symmetricCartans, it suffices to understand the foldings of the connected ones, and forthis it suffices to enumerate the non-trivial automorphisms:

• The only nontrivial automorphism of An is the reflection sending the jthnode to the (n+ 1− j)th node. But this automorphism pairs two verticesconnected by an edge when n is even. When n = 2k+1 is odd, the foldingalong this automorphism is Bk+1.

• The automorphism group of D4 is the symmetric group S3, which hastwo subgroups: the group of order 2 and the group of order 3. But thelatter acts transitively on the three legs of D4, so folding along the group

180

of order 3 or along the full S3 yield the same orbits, and hence the samediagram G2. Folding along the group of order 2 yields the diagram C3.

• For n ≥ 5, the only nontrivial automorphism of Dn permutes the twoshort legs, and the folding is Cn−1

• The group of order 2 acts on E6; the folding of this diagram is an F4.

• E7 and E8 have no nontrivial automorphisms.

**

(c) Verify that every non-symmetric finite type Cartan matrix is obtained by folding froma unique symmetric finite type Cartan matrix.

**We use the above list in the backwards direction: Bn is the folding of A2n+1,Cn the folding of Dn+1, G2 from D4 and F4 from E5.**

6. An indecomposable symmetrizable Cartan matrix A is said to be of affine type if det(A) = 0and all the proper principal minors of A are positive.

(a) Classify the affine Cartan matrices.

**A Dynkin diagram is affine if its determinant is 0 but every proper subdi-agram is the disjoint union of finite-type Dynkin diagrams. Thus every com-ponent but one is of finite-type, and one connected component is affine, sowe need only classify the connected affine diagrams. So we have a connecteddiagram such that the removal of any vertex yields of finite-type diagram.Then the following are forbidden as proper subdiagrams, but allowed as theentire diagram:

i. • ••

••

ii. • •4 // . Higher-degree edges are strictly forbidden.

iii. • • • • •ks , • • • • •+3

iv. • • •_jt , • • •_*4

v. • • • • • • ••

, • • • • • ••

•, • • • • •• •

vi. Any loop of all single edges. Any other loop is strictly forbidden.

vii. The strings • • . . . • •ks +3 , • • . . . • •ks ks , • • . . . • •+3 +3 . A diagramwith two edges, one of degree at least two and the other of degree greaterthan two, is strictly forbidden.

viii. Strings • • . . . • ••

+3 , • • . . . • ••

ks , and • • . . . • •• •

181

Any other connected diagram is either of finite type or contains one of these,so the above is the entire list. **

(b) Show that every non-symmetric affine Cartan matrix is a folding, as in the previousproblem, of a symmetric one.

**We unfold any non-symmetric Cartan matrix to a symmetric one. Let A′

be the folding of A such that detA′ = 0. Then there is a vector ~c =∑cjIj such

that A′~c = 0. But let ~b =∑biαi, where bi = cj if αi ∈ Ij. Then A~b = 0.

So the unfolding of an affine matrix has determinant 0. It need not be affine,though, but by inspection every non-symmetric matrix on the above list hasa symmetric unfolding on the list. **

(c) Let h be a vector space, αi ∈ h∗ and α∨i ∈ h vectors such that A is the matrix 〈αj , α∨i 〉.Assume that this realization is non-degenerate in the sense that the vectors αi are linearlyindependent. Define the affine Weyl group W to be generated by the reflections sαi , asusual. Show that W is isomorphic to the semidirect product W0 nQ where Q and W0

are the root lattice and Weyl group of a unique finite root system, and that every suchW0 nQ occurs as an affine Weyl group.

(d) Show that the affine and finite root systems related as in (c) have the property that theaffine Dynkin diagram is obtained by adding a node to the finite one, in a unique way ifthe finite Cartan matrix is symmetric.

7. Work out the root systems of the orthogonal Lie algebras so(m,C) explicitly, thereby verifyingthat they correspond to the Dynkin diagrams Bn if m = 2n + 1, or Dn if m = 2n. Deducethe isomorphisms so(4,C) ∼= sl(2,C)× sl(2,C), so(5,C) ∼= sp(4,C), and so(6,C) ∼= sl(4,C).

**We take the inverse-diagonal form to be the one preserved by so(m,C), and writeXR for the reflection of a matrix X across the inverse-diagonal, and so(m,C) ={X ∈ gl(m,C) s.t. X = −XR}. Then a basis of so(m,C) are the matrices Xij =Eji − E

m+1−im+1−j for i+ j ≤ m, i, j ≥ 1, where Eji is the matrix that is all 0s except for

a 1 in the (i, j)th spot. The diagonal matrices form the Cartan subalgebra h; abasis is Hi = Xii = Eii − Em+1−i,m+1−i for 1 ≤ i ≤

⌊m2

⌋. Then, using the fact that

[Eii , Elk] = (δki − δli)Elk, we see that

[Hi, Xkl] = (δki − δli)Elk − (δm+1−li − δm+1−k

i )Em+1−km+1−l

− (δkm+1−i − δlm+1−i)Elk + (δm+1−l

m+1−i − δm+1−km+1−i )Em+1−k

m+1−l (PS5.25)

=Äδki − δli + δm+1−l

i − δm+1−ki

äXkl (PS5.26)

and the Xkl diagonalize the action ad : h y so(m,C). We remark that since k+l ≤ mand 2i ≤ m, at most one of the δs in (PS5.26) is non-zero. Let αkl = δki −δli+δ

m+1−li −

δm+1−ki , so that [Hi, Xkl] = αkl(Hi)Xkl. Then as k, l range over k 6= l, k+ l ≤ m, we get

all m(m−1)/2−bm/2c = dm(m− 2)/2e roots. We have αkl = −αlk, and αjk +αkl = αjlif all three of these are defined.

182

Let’s write zi ∈ h∗, 1 ≤ i ≤ bm/2c for the dual basis to Hi, so that zi(Hj) = δij.Then the αkl come in a few types (since αkl = −αlk, we assume that k < l).If k < l ≤ bm/2c, then αkl = zk − zl. If k ≤ bm/2c and m + 1 − l ≤ bm/2c (i.e.l > dm/2e), then αkl = zk + zm+1−l. There is one more possibility, when m is odd:l = dm/2e = bm/2c+ 1; in this case, αkl = zk.

For our notion of positive root, we dot each root αkl against a vector H =∑ziHi ∈ h

such that z1 > z2 > · · · > zbm/2c > 0. Then the positive roots are exactly thosewith k < l. When m is odd, the positive cone is generated by the simple rootsαj

def= αj,j+1 for j = 1, . . . , bm/2c: i.e. αj = zj−zj+1 when j < bm/2c and αbm/2c = zbm/2c.Then the coroots are α∨j = Hj−Hj+1 and αbm/2c = 2Hbm/2c, and we get a root systemof type Bbm/2c:

so(2n+ 1) = • • . . . • •α1 α2 αn−1 αn

+3 (PS5.27)

When m is even, the simple roots still include the roots αjdef= αj,j+1 = zj − zj+1

for j = 1, . . . , bm/2c − 1. But now αm/2,m/2+1 is no longer a root, and on the other

hand αm/2def= αm/2−1,m/2+1 = zm/2−1 +zm/2 is a simple root. The simple co-roots are

α∨j = Hj −Hj+1 for j < m/2 and α∨m/2 = Hm/2−1 +Hm/2, and we get a root system oftype Dm/2:

so(2n) = • • . . . • •

•

α1 α2 αn−2αn−1

αn

(PS5.28)

In any case, we have

so(4) = D2 =•• = A1 tA1 = sl(2)× sl(2) (PS5.29)

so(5) = B2 = • •+3 = C2 = sp(4) (PS5.30)

so(6) = D3 =•• • = A3 = sl(4) (PS5.31)

**

8. Show that the Weyl group of type Bn or Cn (they are the same because these two rootsystems are dual to each other) is the group Sn n (Z/2Z)n of signed permutations, and thatthe Weyl group of type Dn is its subgroup of index two consisting of signed permutationswith an even number of sign changes, i.e., the semidirect factor (Z/2Z)n is replaced by thekernel of Sn-invariant summation homomorphism (Z/2Z)n → Z/2Z

**The Weyl group of a root system is generated by the reflections sα : λ 7→λ−〈λ, α∨〉α for α a root. Then sα = s−α, and so we need only consider the positive

183

roots. The reflections act on the root system, but they also just act (linearly) onh∗, and we will calculate the action on the basic vectors zk.

As calculated in the previous problem, the positive roots of Bn are zi − zj andzi + zj for 1 ≤ i < j ≤ n and zi for 1 ≤ i ≤ n, with co-roots (zi ± zj)∨ = Hi ±Hj andz∨i = 2Hi. Then we have an easy calculation:

szi±zjzk =

zk, i, j 6= k±zj , i = k±zi, j = k

(PS5.32)

szizk =®zk, i 6= k−zk, i = k

(PS5.33)

In particular, the Weyl group of Bn permutes the basic vectors zk up to sign. Itincludes all permutations up to sign — szi±zj acts projectively as the transposition(i, j) — and includes all sign changes, hence must be Sn n ({±1})×n.

The positive roots of Dn, on the other hand, are just {zi ± zj : 1 ≤ i < j ≤ n}, with(zi ± zj)∨ = Hi ±Hj. So the Weyl group is a subgroup of Sn n ({±1})×n. Rewriting(PS5.32), we have

szi±zj =

zk 7→ zk, k 6= i, jzi 7→ ±zjzj 7→ ±zi

(PS5.34)

Thus the Weyl group includes all permutations of the basic vectors, and, by com-posing szi+zj ◦ szi−zj , all the sign-switches in two coordinates. Hence the Weyl

group is Snnker[{±1}×n

q

−→ {±1}], where

q

: {±1}×n → {±1} is the product homo-morphism. **

9. Let (h, R,R∨) be a finite root system, ∆ = {α1, . . . , αn} the set of simple roots with respectto a choice of positive roots R+, si = sαi the corresponding generators of the Weyl group W .Given w ∈W , let l(w) denote the minimum length of an expression for w as a product of thegenerators si.

(a) If w = si1 . . . sir and w(αj) ∈ R−, show that for some k we have αik = sik+1. . . sir(αj),

and hence siksik+1. . . sir = sik+1

. . . sirsj . Deduce that l(wsj) = l(w)− 1 if w(αj) ∈ R−.

**The set R+r{αi} is fixed by all the reflections sj for i 6= j. Thus si1 . . . sir(αj) ∈R+ unless sik+1

. . . sirαj = αik for some k, so that sik can move αk into R−.Then siksik+1

. . . sirαj = sikαik = −αik = −sik+1. . . sirαj = sik+1

. . . sir(−αj) =sik+1

. . . sirsjαj. On the other hand, these reflections are rigid with respectto the right dot product, so sik+1

. . . sir must take the hyperplane perpendic-ular to αj to the hyperplane perpendicular to αik . But on these hyperplanessj and sik respectively act trivially, so on these hyperplanes siksik+1

. . . sir actsas sik+1

. . . sirsj. But then these reflections agree on every vector, since wecan break any vector into components parallel to and perpendicular to αj.

184

Thus wsj = si1 . . . sik−1siksik+1

. . . sirsj = si1 . . . sik−1siksiksik+1

. . . sir = si1 . . . sik−1sik+1

. . . sir .So l(wsj) ≤ l(w) − 1. Conversely, l(w) = l(wsjsj) ≤ l(wsj) + 1, and so we haveequality. **

(b) Using the fact that the conclusion of (a) also holds for v = wsj , deduce that l(wsj) =l(w) + 1 if w(αj) 6∈ R−.

**If wαj ∈ R+, then wsjαj = −wαj ∈ R−, and so by (a) l(w) = l(wsjsj) =l(wsj)− 1. **

(c) Conclude that l(w) = |w(R+) ∪ R−| for all w ∈ W . Characterize l(w) in more explicitterms in the case of the Weyl groups of type A and B/C.

**I assume the formula should read “l(w) = |w(R+) ∩ R−|”. Let r = l(w) andw = si1 . . . sir be a minimal-length factorization of w into simple reflections,and let w1 = wsir . Then l(w1) = r − 1, and so wαir ∈ R−, and w1αir 6∈ R−. Byinduction, l(w1) = |w1(R+)∩R−|. Since sir(R+r{αir}) = R+r{αir}, we see that

|w(R+) ∩R−| = |w(R+ r {αir}) ∩R−|+ |w({αir}) ∩R−| (PS5.35)= |w1sir(R+ r {αir}) ∩R−|+ 1 (PS5.36)= |w1(R+ r {αir}) ∩R−|+ 1 (PS5.37)= l(w1) + 1 = l(w) (PS5.38)

The roots of type An are {ei − ej : 1 ≤ i 6= j ≤ n + 1}, where the simpleroots are when j = i + 1. The positive roots are when i < j, and the Weylgroup is Sn+1 permuting the basis vectors e1, . . . , en+1. So we can canonicallyidentify w ∈ W (An) with a permutation of the numbers 1, . . . , n + 1, and l(w)is the number of pairs of a larger number appearing earlier in the list thana smaller number.

The roots of type Bn, by problem 7, are of the form ±ei ± ej for 1 ≤ i 6= j ≤ nand ±ei for 1 ≤ i ≤ n. The positive roots are ei − ej when i < j, ei + ej,and ei. By problem 8, W (Bn) consists of all signed permutations of the ei;i.e. lists like (2,−4,−3,−5, 1) ∈ W (B5). Then contributing to l(w) are (i) eachminus sign, (ii) each pair i < j that appear as (. . . , j, . . . , i, . . . ), (iii) each pairi < j that appear as (. . . ,−j, . . . , i, . . . ), (iv) each pair i < j that appear as(. . . ,−j, . . . ,−i, . . . ), and (v) each pair i < j that appear as (. . . ,−i, . . . ,−j, . . . ).For example, l(2,−4,−3,−5, 1) = 3 + 1 + 3 + 1 + 2 = 10. **

(d) Assuming that h is over R, show that the dominant cone X = {λ ∈ h : 〈λ, α∨i 〉 ≥0 for all i} is a fundamental domain for W , i.e., every vector in h has a unique elementof X in its W orbit.

**Let w = si1 . . . sir . Let ~ı be the string i1, . . . , ir. A substring ~ = j1, . . . , jtis any string formed from ~ı by deleting letters, including the empty string;

185

write ~ ⊆~ı if ~ is a substring of ~ı. Then

wλ =∑~⊆~ı

(−1)t〈λ, α∨jt〉〈αjt−1 , α∨jt−2〉 . . . 〈αj2 , α∨j1〉αj1 (PS5.39)

where t is the length of ~, and the summand corresponding to the emptystring is just λ itself. There are 2r summands. When w 6= e, the summandscan be sorted by whether ~ contains the initial i1 (necessarily as j1), and thenpaired, one from each pile:

wλ =∑~⊆~ıi1 6∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉αj1

+∑~⊆~ıi1∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉αj1 (PS5.40)

=∑~⊆~ıi1 6∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉αj1

+∑~⊆~ıi1 6∈~

(−1)t+1〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉〈αj1 , α

∨i1〉αi1 (PS5.41)

=∑~⊆~ıi1 6∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉

Äαj1 − 〈αj1 , α∨i1〉αi1

ä(PS5.42)

Thus, bracketing with α∨i1,

〈wλ, α∨i1〉 =∑~⊆~ıi1 6∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉

Ä〈αj1 , α∨i1〉 − 〈αj1 , α

∨i1〉〈αi1 , α

∨i1〉ä

(PS5.43)

=∑~⊆~ıi1 6∈~

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αj2 , α∨j1〉〈αj1 , α

∨i1〉(1− 2) (PS5.44)

since 〈αi, α∨i 〉 = 2 for any i. If λ ∈ X, then 〈λ, α∨j1〉 ≥ 0, and if jn 6= jn+1 forany n, then 〈αjn+1 , α

∨jn〉 ≤ 0. Thus each summand is (−1)t times t non-negative

numbers times −1, so each summand is negative. Or, at least, any summandwithout a repeated index is negative: if jn 6= jn+1 for each jn ∈ ~, then the ~thsummand is negative. The problem occurs only in those summands for whichjn = jn+1 for some n, and then only when this happens an odd number oftimes. But let ~ be a substring of~ıri1 for which jn = jn+1. Then define ~(n) and~(n+1) to be the substrings of ~ formed by removing the nth or (n+1)th letters,respectively; of course, they are the same string and both are substrings of

186

~ı. Letting t be the length of ~, these three strings contribute to the sum asfollows, using the fact that jn = jn+1:

(−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αjn+2 , α

∨jn+1〉〈αjn+1 , α

∨jn〉〈αjn , α

∨jn−1〉 . . . 〈αj2 , α∨j1〉〈αj1 , α

∨i1〉(−1)

+ (−1)t−1〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αjn+2 , α

∨jn+1〉〈αjn+1 , α

∨jn−1〉 . . . 〈αj2 , α∨j1〉〈αj1 , α

∨i1〉(−1)

+ (−1)t−1〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αjn+2 , α

∨jn〉〈αjn , α

∨jn−1〉 . . . 〈αj2 , α∨j1〉〈αj1 , α

∨i1〉(−1)

= (−1)t〈λ, α∨jt〉〈αjt , α∨jt−1〉 . . . 〈αjn+2 , α

∨jn+1〉(2− 1− 1)〈αjn , α∨jn−1

〉 . . . 〈αj2 , α∨j1〉〈αj1 , α∨i1〉(−1)

= 0 (PS5.45)

So any string with a repeated index cancels exactly with a shorter string.

A problem occurs when jn = i1, as in this case the various terms don’t cancelexactly. Let m be the lowest number m > 1 so that im = i1. The onlyproblematic summands are when ~ ⊆~ır i1 has j1 = im. Then the sum includesstrings of the form

(−1)t〈λ, α∨jt〉 . . . 〈λ, α∨j2〉〈αj2 , α

∨j1〉〈αj1 , α

∨i1〉

+ (−1)t−1〈λ, α∨jt〉 . . . 〈λ, α∨j2〉〈αj2 , α

∨i1〉

+∑

1<n<m

(−1)t〈λ, α∨jt〉 . . . 〈λ, α∨j2〉〈αj2 , α

∨in〉〈αin , α

∨i1〉 (PS5.46)

= (−1)t〈λ, α∨jt〉 . . . 〈λ, α∨j2〉(

2− 1 +∑

1<n<m

〈αin , α∨i1〉)

(PS5.47)

Then, since 〈αi, α∨j 〉 is always an integer, non-positive unless i = j, the term inparentheses in (PS5.47) is non-positive unless 〈αin , α∨i1〉 = 0 for every 1 < n <m, and the rest of the terms in the whole sum are good (if ~ı contains anotherterm equal to i1 — say m2 is the smallest number greater than m = m1 so thatim2 = im1, then we repeat the trick with a sum over m < n < m2). But thenwe can commute the reflections sin past si1 until si1 is next to sim, and theycan cancel each other. So our expression for w was not of minimal length.

Lastly, we need only treat the case when 〈wλ, α∨i1〉 = 0. But then si1wλ = wλ,and si1w is strictly shorter than w, assuming we wrote w as a minimal-lengthproduct of sis. So we can proceed by induction, and conclude that wλ 6∈ X ifwλ 6= λ.

In any case, we see that if λ ∈ X and w 6= e, then wλ 6∈ X. Conversely, wedefine the height of λ to be the sum of the coefficients ci when we expandλ in terms of the basis λ =

∑ciαi. If λ 6∈ X, then for some i, 〈λ, α∨i 〉 < 0;

then siλ = λ−〈λ, α∨i 〉αi has an increased height, since no coefficient cj changesexcept for ci 7→ ci+|〈λ, α∨i 〉|. Since W is finite (or, anyway, a closed subgroup ofSO(dim h), hence compact), the maximum height of the orbit of λ is achieved,but it can only be achieved in X.

187

This proves that X is a fundamental domain of W y h. **

(e) Deduce that |W | is equal to the number of connected regions into which h is separatedby the removal of all the root hyperplanes 〈λ, α∨〉, α∨ ∈ R∨.

**The set h6=0 = hr root hyperplanes = {λ ∈ h s.t. 〈λ, α∨i 〉 6= 0∀i} is fixed underthe W action. Then each orbit in W y h 6=0 intersection X exactly once,still, and so X ∩ h6=0 is a fundamental domain. By symmetrizability, eachhyperplane is honestly perpendicular to the corresponding root, and since theαi are simple for a system of positive roots, X∩h 6=0 = {λ ∈ h s.t. 〈λ, α∨i 〉 > 0 ∀i},which is connected.

One can trace through the proof of (d) and conclude that wλ = λ only ifsi1wλ = wλ and so 〈wλ, α∨i1〉 = 0, for w = si1 . . . sir a minimal factorization. Butthen 〈λ, α∨i1〉 = 0, and so λ 6∈ h6=0. So w1X ∩ w2X = ∅ unless w1 = w2, and onthe other hand h 6=0 =

⋃w∈W wX. So |W | is the number of connected regions

in h6=0. **

**Incidentally, I’m not sure what parts (d) and (e) have to do with parts (a-c).**

10. Let h1, . . . , hr be linear forms in variables x1, . . . , xn with integer coefficients. Let Fq denotethe finite field with q = pe elements. Prove that except in a finite number of “bad” charac-teristics p, the number of vectors v ∈ Fnq such that hi(v) = 0 for all i is given for all q by apolynomial χ(q) in q with integer coefficients, and that (−1)nχ(−1) is equal to the numberof connected regions into which Rn is separated by the removal of all the hyperplanes hi = 0.

Pick your favorite finite root system and verify that in the case where the hi are the roothyperplanes, the polynomial χ(q) factors as (q − e1) . . . (q − en) for some positive integers eicalled the exponents of the root system. In particular, verify that the sum of the exponentsis the number of positive roots, and that (by Problem 9(e)) the order of the Weyl group is∏i(1 + ei)

11. The height of a positive root α is the sum of the coefficients ci in its expansion α =∑i ciαi

on the basis of simple roots.

Pick your favorite root system and verify that for each k ≥ 1, the number of roots of heightk is equal to the number of the exponents ei in Problem 10 for which ei ≥ k.

12. Pick your favorite root system and verify that if h denotes the height of the highest rootplus one, then the number of roots is equal to h times the rank. This number h is calledthe Coxeter number. Verify that, moreover, the multiset of exponents (see Problem 10) isinvariant with respect to the symmetry ei 7→ h− ei.

**My favorite root systems are F4 and E6. In F4, the highest root is 2 3 4 2+3 ,so h = 12 and there are 12 × 4 = 48 non-zero roots. In E6 the highest root is

1 2 32

2 1, h = 12, and there are 72 non-zero roots.

188

By the previous problem, the number of exponents equal to k is the number ofroots of height k minus the number of roots of height k + 1. In F4, for example,using problem 16 below, we can count the number of roots at a given height, andhence the number of exponents:

k # roots at height k # exponents = k

1 4 12 3 03 3 04 3 05 3 16 2 07 2 18 1 09 1 0

10 1 011 1 112 0 0

Sure enough, the multiset is appropriately symmetric. **

13. A Coxeter element in the Weyl group W is the product of all the simple reflections, once each,in any order. Prove that a Coxeter element is unique up to conjugacy. Pick your favoriteroot system and verify that the order of a Coxeter element is equal to the Coxeter number(see Problem 12).

**It suffices to be able to switch the places of two adjacent reflections in a Coxeterelement. Let α1, α2 be two nodes in the Dynkin diagram, and . . . s1s2 . . . a Coxeterelement. If α1 connect to α2, then . . . s1s2 · · · = . . . s2s1 . . . . If α1 is connected toα2 and nothing else, then conjugating by s1 gives s1(. . . s1s2 . . . )s1 = . . . s2s1 . . . .Similarly, we can conjugate by s2 if α2 connects to only α1. Thus α1 and α2 eachconnect to another vertex, and α1 and α2 cannot both be trivalent. So withoutloss of generality we can assume that α2 connects only to α1 and α3: our diagramis . . . α1 α2 α3 . . ., where possibly the edges are double- or triple-arrows.Conjugation includes cyclic permutations, so we can assume that our Coxeterelement is . . . s3 . . . s1s2 . . . . Then every other reflection commutes with s2, andconjugating by s2 gives s2(. . . s3 . . . s1s2 . . . )s2 = . . . s2s3 . . . s1 · · · = . . . s2s3s2 . . . s2s1 . . . .So we can push an s1 past an s2 if and only if we can push an s2 past an s3.

We repeat the story. If α3 is at the end of its chain, then we can push s2 pasts3 at the cost of conjugation. Otherwise it links to an α4, and we continue. Theonly obstruction is if we eventually run into a trivalent vertex. But there can beat most one trivalent vertex in any connected component of a Dynkin diagram,so we can move either in the α1 direction or the α2 direction until we get to theend of a leg. An induction argument on the distance of a vertex to the end of its

189

leg would be the right way to write this up formally. **

14. The fundamental weights λi are defined to be the basis of the weight lattice P dual to thebasis of simple coroots in Q∨, i.e., 〈λi, α∨j 〉 = δij .

(a) Prove that the stabilizer in W of λi is the Weyl group of the root system whose Dynkindiagram is obtained by deleting node i of the original Dynkin diagram.

**Since sαλ = λ − 〈λ, α∨〉α∨, a reflection sα fixes a weight λ if and only if〈λ, α∨〉 = 0, i.e. if and only if λ and α are orthogonal. Let H be a Dynkin dia-gram, αi a simple root, λi its fundamental weight, and Hi = Hrαi the diagramformed by deleting the node αi from H. Then the reflections correspondingto roots in Hi fix λi; these reflections generate the subgroup W (Hi) ⊆ W (H),and so W (Hi) fixes λi.

However, I’m not sure why W (Hi) is the entire stabilizing subgroup. **

(b) Show that each of the root systems E6, E7, and E8 has the property that its highest rootis a fundamental weight. Deduce that the order of the Weyl group W (Ek) in each caseis equal to the number of roots times the order of the Weyl group W (Ek−1), or W (D5)for k = 6. Use this to calculate the orders of these Weyl groups.

**Computing a fundamental weight is easy given computers: we work in thesimple-root basis, and then the fundamental weights are the columns of theinverse matrix A−1, where A is the Cartan. For

En = α1 . . . αn−4 αn−3

αn

αn−2 αn−1,

190

these are:

2 −1 0 0 0 0−1 2 −1 0 0 00 −1 2 −1 0 −10 0 −1 2 −1 00 0 0 −1 2 00 0 −1 0 0 2

−1

=

113 12

3 2 113

23 1

123 31

3 4 223 11

3 22 4 6 4 2 3

113 22

3 4 313 12

3 223 11

3 2 123 11

3 11 2 3 2 1 2

(PS5.48)

2 −1 0 0 0 0 0−1 2 −1 0 0 0 00 −1 2 −1 0 0 00 0 −1 2 −1 0 −10 0 0 −1 2 −1 00 0 0 0 −1 2 00 0 0 −1 0 0 2

−1

=

112 2 21

2 3 2 1 112

2 4 5 6 4 2 321

2 5 712 9 6 3 41

23 6 9 12 8 4 62 4 6 8 6 3 41 2 3 4 3 2 2

112 3 41

2 6 4 2 312

(PS5.49)

2 −1 0 0 0 0 0 0−1 2 −1 0 0 0 0 00 −1 2 −1 0 0 0 00 0 −1 2 −1 0 0 00 0 0 −1 2 −1 0 −10 0 0 0 −1 2 −1 00 0 0 0 0 −1 2 00 0 0 0 −1 0 0 2

−1

=

2 3 4 5 6 4 2 33 6 8 10 12 8 4 64 8 12 15 18 12 6 95 10 15 20 24 16 8 126 12 18 24 30 20 10 154 8 12 16 20 14 7 102 4 6 8 10 7 4 53 6 9 12 15 10 5 8

(PS5.50)

A column of all integers corresponds to a fundamental weight in that is alsoin Q. In each case, the matrix contains a root:

E6 : λ6 = 1 2 3

2

2 1 = s6s3s2s4s1s3s5s2s4s6 (α1) (PS5.51)

E7 : λ6 = 1 2 3 4

2

3 2 = s6s5s4s3s2s1

Ö0 1 2 3

2

2 1

è(PS5.52)

E8 : λ1 = 2 3 4 5 6

3

4 2

= s1s2s3s4s5s8s6s5s4s3s2s1

Ö0 1 2 3 4

2

3 2

è(PS5.53)

191

(We use the previous line to continue the simplification to a simple root.)

As is well known, the size of any group equals the size of an orbit of an actionof the group times the size of the stabilizing subgroup of any element in theorbit, and the Weyl group of En acts transitively on the non-zero roots.The diagram E6 r α6 is an A5; E6 has 72 non-zero roots, and so |W (E6) =|W (A5)| × 72 = 5! × 72 = 8640. E7 has 126 non-zero roots and E6 r α6 = D6, so|W (E7)| = |D5|×126 = 241 920. By my calculations, only E8 has a root that is afundamental weight at a node the removal of which leaves the diagram in theE series. With 240 roots for E8, we have |W (E8)| = |W (E7)| × 240 = 58 060 800.**

15. Let e1, . . . , e8 be the usual orthonormal basis of coordinate vectors in Euclidean space R8. Theroot system of type E8 can be realized in R8 with simple roots αi = ei − ei+1 for i = 1, . . . , 7and

α8 =Å−1

2,−1

2,−1

2,12,12,12,12,12

ã.

Show that the root lattice Q is equal to the weight lattice P , and that in this realization, Qconsists of all vectors β ∈ Z8 such that

∑i βi is even and all vectors β ∈

Ä12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12

ä+

Z8 such that∑i βi is odd. Show that the root system consists of all vectors of squared length

2 in Q, namely, the vectors ±ei ± ej for i < j, and all vectors with coordinates ±12 and an

odd number of coordinates with each sign.

**Let p be a vector in the weight lattice P . Then 〈p, α∨j 〉 = Ap, where A is theCartan matrix

A =

2 −1−1 2 −1

−1 2 −1 −1−1 2 −1

−1 2 −1−1 2 −1

−1 2−1 2

(PS5.54)

in terms of the αi-basis — then Ap is a vector of all integers, by definition. ButdetA = 1, so p is a vector of all integers. Indeed, the computer calculates:

A−1 =

4 7 10 8 6 4 2 57 14 20 16 12 8 4 1010 20 30 24 18 12 6 158 16 24 20 15 10 5 126 12 18 15 12 8 4 94 8 12 10 8 6 3 62 4 6 5 4 3 2 35 10 15 12 9 6 3 8

(PS5.55)

192

In any case, then p ∈ Q.

We now work in the representation αi = ei−ei+1 for 1 ≤ i ≤ 7 and α8 =Ä−1

2 ,−12 ,−

12 ,

12 ,

12 ,

12 ,

12 ,

12

ä.

Let X be the sublattice of P = Q generated by {αi : 1 ≤ i ≤ 7}∪{2α8}. Then X ⊆ Z8,and if x = (x1, x2, . . . , x8) ∈ X, then

∑xi is even. Conversely, if x ∈ Z8 with

∑xi

even, then x is in the lattice generated by the αi, i ≤ 7, and 2α8, since

1 0 0 0 0 0 0 −1−1 1 0 0 0 0 0 −10 −1 1 0 0 0 0 −10 0 −1 1 0 0 0 10 0 0 −1 1 0 0 10 0 0 0 −1 1 0 10 0 0 0 0 −1 1 10 0 0 0 0 0 −1 1

−1

=

112

12

12

12

12

12

12

12

2 2 1 1 1 1 1 121

2 212 21

2 112 11

2 112 11

2 112

2 2 2 2 1 1 1 111

2 112 11

2 112 11

212

12

12

1 1 1 1 1 1 0 012

12

12

12

12

12

12 −1

212

12

12

12

12

12

12

12

(PS5.56)

In particular, each row on the right is composed entirely either of integers or ofhalf-more-than integers. So if

∑xi is even, then multiplying x by the matrix on

the right yields integer coefficients in the {α1, . . . , α7, 2α8}-basis.

In any case, X is clearly index-2 in Q = P , and Q = X ∪ (α8 + X). But thesum of the coefficients of α8 =

Ä−1

2 ,−12 ,−

12 ,

12 ,

12 ,

12 ,

12 ,

12

äis odd, so α8 + X ⊆ {β ∈Ä

12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12

ä+Z8 :

∑βi is odd}, and conversely if β ∈

Ä12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12

ä+Z8

and∑βi is odd, then β − α8 is a vector of integers whose sum is even.

As for understanding the full root system, we observe that all the simple rootshave square-length 2, and as always the Weyl group acts on the root systemtransitively. If 1 ≤ i < j ≤ 8, then ei − ej = sisi+1 . . . sj−2αj−1, and ej − ei =sisi+1 . . . sj−2sj−1αj−1. Moreover,

e7 + e8 = s8s3s2s4s1s5s3s2s4s8α3 (PS5.57)

and ei+ej = sjsj+1 . . . s7sisi+1 . . . s6(e7+e8). For the negatives, replace α3 in (PS5.57)by s3α3. Thus, any vector in X from the previous paragraph with length-squaredequal to 2 is a root.

To construct the roots in X+α8, we use the fact that for j ≤ 7, sj switches ej withej+1 and leaves the rest of the ei fixed. Indeed, s1, . . . , s7 generate the symmetricgroup S8 on 8 letters {e1, . . . , e8}. So we can achieve any vector in {±1

2}8 with

exactly three −12s be acting on α8 by the appropriate element of S8. Similarly,

any element with exactly five −12s is in the S8-orbit of −α8 = s8α8. On the other

hand,Ä−1

2 ,12 ,

12 ,

12 ,

12 ,

12 ,

12 ,

12

ä= α1 + 3α2 + 5α3 + 4α4 + 3α5 + 2α6 + α7 + 3α8 = s8(e2 + e3),

and by acting by S8 we get all length-√

2 vectors in X + α8 with exactly one −12 .

The vectors with seven −12s are just the corresponding negative roots.

193

We have exhibited that all vectors in Q with square-length 2 are roots. Thereare, of course, 4×

(82

)+(81

)+(83

)+(85

)+(87

)= 240 of these, and so e8 is 240 + 8 = 248-

dimensional. **

16. Show that the root system of type F4 has 24 long roots and 24 short roots, and that the rootsof each length form a root system of type D4. Show that the highest root and the highestshort root are the fundamental weights at the end nodes of the diagram. Then use Problem14(a) to calculate the order of the Weyl group W (F4). Show that W (F4) acts on the set ofshort (resp. long roots) as the semidirect product S3 nW (D4), where the symmetric groupS3 on three letters acts on W (D4) as the automorphism group of its Dynkin diagram.

**We list all the positive roots:

• 1 1 1 1 1 1 1 2 1

•��

1 1 2 2 2 3 3 1 1 1 1

• 2 2 2 4 4 4 2 2 2

• 2 2 2 2 2 2

• 1 1 1 1 1 1

•��

1 1 1 1 1 1 2 2 2

• 1 1 1 1 1 1 2 2 2 3 3

• 1 1 1 1 1 1 1 1 2

Taking the extremal roots among the positive short or long roots gives systemsof type D4, best expressed in a big diagram:

0001

0010

0110

1110

111111

1000

0100

0120

0122

111111

194

The highest root and highest short root are (2, 3, 4, 2) and (1, 2, 3, 2), respectively; itis immediate to check that these are the fundamental weights at the end nodes.Then problem 14(a) provides that the stabilizer in W (F4) of each of these isW (B3) = W (C3), which has size 6 × 8 = 48 by problem 8 above. Since W (F4) actstransitively on the roots of a given length, we have |W (F4)| = 24× 48 = 1152.

Since F4 contains D4 as the system of short (long) roots, and since swα = wsαw−1

for any w ∈ W (F4) and W (F4) preserves the length of a root, W (D4) is a normalsubgroup of W (F4). Labeling the roots of F4 as α1 α2 α3 α4+3 , we see that sα1

and sα2 act on the D4 of short roots as two of the reflections (hence generate)in the symmetric group S3. Indeed, it’s true in general that (sα1sα2)3 = id, andby comparing sizes of groups — |W (D4)| = 192 by problem 8 — we see thatW (F4) = S3 nW (D4). **

17. Pick your favorite root system and verify that the generating function W (t) =∑w∈W tl(w) is

equal to∏i (1 + t+ · · ·+ tei), where ei are the exponents as in Problem 10.

18. Let S be the subring of W -invariant elements in the ring of polynomial functions on h. Pickyour favorite root system and verify that S is a polynomial ring generated by homogeneousgenerators of degrees ei + 1, where ei are the exponents as in Problem 10.


1. Show that the simple complex Lie algebra g with root system G2 has a 7-dimensional matrixrepresentation with the generators shown below.

e1 =

0 0 0 0 0 0 00 0 1 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 1 00 0 0 0 0 0 00 0 0 0 0 0 0

f1 =

0 0 0 0 0 0 00 0 0 0 0 0 00 1 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 1 0 00 0 0 0 0 0 0

e2 =

0 1 0 0 0 0 00 0 0 0 0 0 00 0 0 2 0 0 00 0 0 0 1 0 00 0 0 0 0 0 00 0 0 0 0 0 10 0 0 0 0 0 0

f2 =

0 0 0 0 0 0 01 0 0 0 0 0 00 0 0 0 0 0 00 0 1 0 0 0 00 0 0 2 0 0 00 0 0 0 0 0 00 0 0 0 0 1 0

(PS6.1)

**Letting Eji be the basic matrix with all 0s but a 1 at the (i, j)th spot, we recall

195

that [Eji , Elk] commute unless i = l or j = k. In particular, [Eji , E

ij ] = Eii − Ejj ,

and so e1 and f1 generate an sl(2) with h1 = E22 − E3

3 + E55 − E6

6 , and [e2, f2] =E1

1 − E22 + 2E3

3 − 2E44 + 2E4

4 − 2E55 + E6

6 − E77 , so we have

h1 =

01−1

01−1

0

h2 =

1−1

20−2

1−1

Recall moreover that [

∑aiE

ii , E

kj ] = (aj − ak)Ekj ; then h1 commutes only with diag-

onal matrices and those of the form

0 0 0 ∗ 0 0 ∗0 0 0 0 ∗ 0 00 0 0 0 0 ∗ 0∗ 0 0 0 0 0 ∗0 ∗ 0 0 0 0 00 0 ∗ 0 0 0 0∗ 0 0 ∗ 0 0 0

whereas h2 commutes only with diagonal matrices and those of the form

0 0 0 0 0 ∗ 00 0 0 0 0 0 ∗0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0∗ 0 0 0 0 0 00 ∗ 0 0 0 0 0

So barring unforeseen diagonal matrices, h = 〈h1, h2〉 is a Cartan. Anyway, we candive ahead and compute the rest of the algebra: [e1, f2] = 0 = [e2, f1],

[e1, e2] =

0 0 −1 0 0 0 00 0 0 2 0 0 00 0 0 0 0 0 00 0 0 0 0 −1 00 0 0 0 0 0 10 0 0 0 0 0 00 0 0 0 0 0 0

[f1, f2] =

0 0 0 0 0 0 00 0 0 0 0 0 01 0 0 0 0 0 00 −1 0 0 0 0 00 0 0 0 0 0 00 0 0 2 0 0 00 0 0 0 −1 0 0

Of course, [hi, ei] = 2ei and [hi, fi] = −2fi, and [h1, e2] = −e2, [h1, f2] = f2, [h2, e1] =−3e1, and [h2, f1] = 3f1, and by Jacobi hi acts diagonally on the basis given by

196

bracket monomials —

if [h, a] = αa and [h, b] = αb, then [h, [a, b]] = (α+ β)[a, b]. (PS6.2)

Then [[e1, e2], fi] = [[e1, fi], e2] + [e1, [e2, fi]] = [h1, e2] or [e1, h2], and [[e1, e2], [f1, f2]] =[[[e1, e2], f1], f2] + [f1, [[e1, e2], f2]] = [[h1, e2], f2] + [f1, [e1, h2]] is also already in the spanof known things. But [e1, [e1, e2]] = 0, since it it strictly upper triangular but[h1, [e1, [e1, e2]]] = 0, and

[e2, [e1, e2]] =

0 0 0 4 0 0 00 0 0 0 −2 0 00 0 0 0 0 −1 00 0 0 0 0 0 10 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0

[f2, [f1, f2]] =

0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 01 0 0 0 0 0 00 −1 0 0 0 0 00 0 −2 0 0 0 00 0 0 4 0 0 0

Again bracketing [e2, [e1, e2]] with fi or [f2, [f1, f2]] with ei does not give new basiselements, by Jacobi. By now I really should stop. It’s obvious we’ll never getmore diagonals, so we have a Lie algebra with a rank-two Cartan which is notone of A1×A1, A2, and B2, and so must be G2, provided the algebra is semisimple.In fact, this algebra is simple: in an ideal, find some h-diagonal entries, whichwe can describe explicitly in terms of the monomial basis and are confined to(off-)diagonals, bracket with the opposite off-diagonals until we get a diagonalmatrix, and bracket with the es and fs to get the generators. On the other hand,we might as well complete the calculation. Mixing es with fs doesn’t make fornew basis vectors. [e1, [e2, [e1, e2]]] = [[e1, e2], [e1, e2]] + [[e1, [e1, e2]], e2] = 0. Lastly,

[e2, [e2, [e1, e2]]] =

0 0 0 0 −6 0 00 0 0 0 0 0 00 0 0 0 0 0 40 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0

[f2, [f2, [f1, f2]]] =

0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 04 0 0 0 0 0 00 0 0 0 0 0 00 0 −6 0 0 0 0

[e1, [e2, [e2, [e1, e2]]]] =

0 0 0 0 0 6 00 0 0 0 0 0 40 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0

[f1, [f2, [f2, [f1, f2]]]] =

0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 04 0 0 0 0 0 00 6 0 0 0 0 0

and [e2, [e2, [e2, [e1, e2]]]] = [f2, [f2, [f2, [f1, f2]]]] = 0, and the algebra is 14-dimensional.

197

In any case, in the h basis, α1 = (2,−1) and α2 = (−3, 2) are two roots of theproposed Lie algebra. The monomials in all es are positive roots, those in all fsthe negatives, and by the addition formula (PS6.2), these are simple roots. Theroots correspond to the monomial basis, of course.

0α1 = e1

α2 = e2

• = [e1, e2]

• = [e2, [e1, e2]]

• = [e2, [e2, [e1, e2]]]

• = [e1, [e2, [e2, [e1, e2]]]]

h∗1

h∗2

//

OO

''OOOOOOO

ddJJJJJJJJJJJJ

Anyway, all this is interesting, but beside the point. The point is to evaluate theCartan, which we did long ago, as soon as we wrote down hi and ei:

〈αi, α∨j 〉 = αi(hj) =Ç

2 −1−3 2

å(PS6.3)

We conclude by redrawing the positive roots with the correct angles. These comefrom writing h1 = (0, 1,−1) and h2 = (1,−1, 2) as vectors in R3, thought of as thediagonal 7× 7 matrices, antisymmetric across the anti-diagonal. Then (h1, h1) = 2,(h2, h2) = 6, and (h1, h2) = −3. Thus up to a rescaling we can take α1 = (1, 0) andα2 = (−3/2,

√3/2):

0 α1

α2 •

•

• •

//

OO

//

ffMMMMMMM

**

2. (a) Show that there is a unique Lie group G over C with Lie algebra of type G2.

**The Cartan matrix has determinant 1. We used the fact already in theprevious problem set that if a root system has a Cartan of determinant 1,then the root and weight lattices match, and hence there is only one complexconnected Lie group integrating the given root system. The justification isas follows. Let {αi} be the simple roots, forming a basis of h = h∗, which weidentify via αi 7→ α∨i . If β =

∑biαi and γ =

∑ciαi are two vectors in h and A is

the Cartan, then by definition the pairing is given by matrix multiplication:〈β, γ∨〉 = ~bTA~c. Thus ~c is in the weight lattice exactly if A~c is a vector of all

198

integers. But if detA has determinant 1, then A−1 is a matrix of all integers(in general, the entries of A−1 are polynomials in the entries of A, which areintegers, divided by detA), and so ~c = A−1A~c is a vector of all integers. Henceif detA = 1, then the only weights are integer combinations of the simpleroots, i.e. are in the root lattice. And if the root and weight lattices agree,then there is only one connected group with the given Lie algebra. **

(b) Find explicit equations of G realized as the algebraic subgroup of GL(7,C) whose Liealgebra is the image of the matrix representation in Problem 1.

**We remark that the LDU decomposition of a matrix is unique if it exists,and if the LDU decomposition of a matrix exists, then it is given by theDoolittle algorithm, which writes the entries of the LDU decomposition ofa matrix X as rational functions in the entries of X. Indeed, given a matrixX, let ∆j1,...,jr

i1,...,ir(X) be the determinant of the minor with rows {i1, . . . , ir} and

columns {j1, . . . , jr}, and ∆i(X) = ∆1,...,i1,...,i(X) the ith leading principal minor.

By convention, ∆j1,...,jri1,...,ir

is antisymmetric in its indices: in particular, if anylower index or any upper index is repeated, then ∆ = 0. We will also adoptthe convention that ∆0 = 1 is the determinant of the 0 × 0 matrix. ThenX = U−DU+, with

(U−)ji (X) =∆1,2,...,j−1,i

1,2,...,j−1,j(X)∆j(X)

, Dji (X) = δji

∆i(X)∆i−1(X)

, and (U+)ji (X) =∆1,2,...,i−1,j

1,2,...,i−1,i (X)∆i(X)

(PS6.4)

So, say we’re interested in G = G2 with g2 = Lie(G2) a subalgebra of M7(C) =gl(7,C) as in problem 1. Then g2 = n−⊕h⊕n+ as always, with n− the span of thebracket-monomials in f1, f2, h = span{h1, h2}, and n+ the span of the bracketmonomials in ei. Then n± are six-dimensional linear spaces subspaces of M7,and so each is the zero set of some collection of 43 linear functionals P± on M7.Let L(X) be the sixth-degree polynomial M7 →M7 : X 7→∑

k=1(1−X)k/k. ThenX ∈ exp(n±) = N± if and only if L(X) ∈ n±, i.e. if and only if P± ◦ L±(X) = 0.

The vector space h consists of diagonal matrices X with Xii = −X8−i

8−i fori = 1, . . . , 4 (in particular, X4

4 = 0), and such that X11 = X2

2 +X33 . So exp(h) = T

consists of diagonal matrices X with XiiX

8−18−i = 1, i = 1, 2, 3, and X4

4 = 1, suchthat XX

1 = X22X

33 . If X ∈ G, then detX = 1, since g consists of traceless

matrices. If X ∈ G has an LDU decomposition, then L(X) ∈ N−, D(X) ∈ T ,and U(X) ∈ N+. In particular, (∆1)2 = ∆3, and

1 = ∆7(X), ∆1 = ∆6, ∆2 = ∆5, and ∆3 = ∆4 (PS6.5)

Conversely, the elements of G with LDU decomposition are a dense subsetof G, but the solution to (PS6.5) is closed in G, and so all of X ∈ G satisfies(PS6.5). We need only state the conditions that U±(X) ∈ N±. Of course,

199

U±(X) is not a polynomial in X. But we take the equations P± ◦L◦U±(X) = 0,put everything over a common denominator in lowest terms (remember thatK[x1, . . . , xn] is a UFD for any n), and take the numerators as our definingpolynomials.

Of course, this description is not very explicit. Indeed, up to possibly makingthe degree of L(X) higher, the previous discussion applies essentially always.For an explicit description of G2, we remember hearing somewhere that G2 yK7 as automorphisms of the octonions. These are the non-associative eight-dimensional unital algebra, defined by the multiplication table

× x1 x2 x3 x4 x5 x6 x7

x1 −1 x3 −x2 x5 −x4 −x7 x6

x2 −x3 −1 x1 x6 x7 −x4 −x5

x3 x2 −x1 −1 x7 −x6 x5 −x4

x4 −x5 −x6 −x7 −1 x1 x2 x3

x5 x4 −x7 x6 −x1 −1 −x3 x2

x6 x7 x4 −x5 −x2 x3 −1 −x1

x7 −x6 x5 x4 −x3 −x2 x1 −1

(PS6.6)

Separating real and imaginary parts, the octonions are defined by the negative-definite symmetric form −δij along with the antisymmetric (2, 1)-tensor µkij :K7 ⊗ K7 → K7 given by the non-diagonal entries in (PS6.6). We can thendefine the automorphism group of the octonians to be those matrices thatpreserve δij and µkij. To preserve δij, a matrix X must satisfy 28 quadraticequations in the coefficients (one for each symmetric pair {i, j}). To preserveµkij, X must satisfy naively another 7×21 equations, these all cubic. However,contracting µkij with δkl gives a totally antisymmetric form µijk : (K×7)⊗3 → K,and it is really this form which must be preserved; this can be written withonly

(73

)= 35 cubic equations. So the automorphism group is cut out of GL(7)

by 28 + 35 = 63 equations. But GL(7) is only 49-dimensional, so some ofthese equations are redundant. Anyway, we write them down without anydifficulty: ∑

m,n

Xmi X

nj δmn = δij and

∑m,n,p

Xmi X

nj X

pkµmnp = µijk (PS6.7)

If δ and µ are the diagonal and off-diagonal parts of (PS6.6), then the differ-ential automorphisms X ∈ Lie(Aut(octonions)) satisfy X +XT = 0 and

Xmi µmjk +Xm

j µimk +Xmk µijm = 0. (PS6.8)

If i = j, then (PS6.8) vanishes by antisymmetry. If µijk = ±1, then (PS6.8)says Xi

i ±Xjj ±Xk

k = 0, but Xmm = 0 since X +XT = 0. Thus there are

(73

)− 7 =

200

28 non-trivial equations in (PS6.5), but only one quarter of these are notredundant, recalling that Xj

i +Xij = 0:

0 = X61 +X2

5 +X34 (PS6.9)

= X71 +X4

2 +X35 (PS6.10)

= X41 +X2

7 +X63 (PS6.11)

= X51 +X6

2 +X73 (PS6.12)

= X21 +X7

4 +X56 (PS6.13)

= X31 +X4

6 +X57 (PS6.14)

= X32 +X4

5 +X67 (PS6.15)

So really the Lie algebra is sliced out by 28 + 7 = 35 equations (28 for δij), andhence is the 14-dimensional g2.

But the generators of g2 given in (PS6.1) do not preserve this presentationof the octonions. Indeed, they do not preserve a symmetric nondegeneratepairing like (a, b) 7→ <(ab) on the octonions. If γij is symmetric and preservedby e1 and f1, then

γ =

∗ 0 0 0 0 ∗ ∗0 0 0 0 −a 0 00 0 0 a 0 0 00 0 a 0 0 0 00 −a 0 0 0 0 0∗ 0 0 0 0 ∗ ∗∗ 0 0 0 0 ∗ ∗

(PS6.16)

But if γij is also preserved by e2, then in particular a = 0.

So something is going screwy here. **

3. Show that the simply connected complex Lie group with Lie algebra so(2n,C) is a doublecover Spin(2n,C) of SO(2n,C), whose center Z has order four. Show that if n is odd, thenZ is cyclic, and there are three connected Lie groups with this Lie algebra: Spin(2n,C),SO(2n,C) and SO(2n,C)/{±I}. If n is even, then Z ∼= (Z/2Z)2, and there are two more Liegroups with the same Lie algebra.

**We have earlier (Problem Set 5, exercise 7) worked out the root system ofso(2n,C). It’s given by the Dynkin diagram

so(2n) = • • . . . • •

•

α1 α2 αn−2αn−1

αn

(PS6.17)

201

and an explicit description of the roots is that αj = zj − zj+1 for j < n andαn = zn−1 + zn. The simple co-roots are α∨i = hi − hi+1 for i < n and α∨n = hn−1 + hn,where {hi} are a basis of h and 〈zj , hi〉 = δij. Probably there ought to be some raisedindices here. Then the root lattice Q consists of those vectors in Zn = Z{zj} ⊆ h∗

such that the sum of the coefficients is even. Let β = (12 , . . . ,

12); then the weight

lattice is P = Zn t (β + Zn). The index [P : Q] = 4 is the size of the center of thesimply-connected Lie group; indeed, this center is P/Q by general theory. Weknow from Problem Set 1, exercise 1(d), that the center of SO(2n) has size 2, soSO(2n) corresponds to a lattice of index-2 between P and Q. But the definingrepresentation SO(2n) y C2n splits into generalized weight spaces with weights±zi, so SO(2n) corresponds to the lattice Zn = Z{zj} between Q and P .

Let X be a lattice Q ( X ( P , and x ∈ X r Q. If n is odd, then for any x ∈ P ,x ∈ Zn exactly when the sum of the coefficients of x is an integer; if x ∈ Zn r Q,then Q ∪ (x+Q) = Zn, and so X = Zn. On the other hand, if x ∈ P r Zn, then thesum of the coefficients of x is either half more or half less than an even integer,and in either case 2x ∈ ZnrQ. So the only lattice between Q and P is Zn. Anotherway to say this is that P/Q = {0, 1/2, 1, 3/2}, and by general nonsense this cyclicgroup is the center of the simply-connected group.

If n is even, then 2β ∈ Q, and so P/Q contains two different elements (the equiv-alence classes of β and of (1, 0, . . . )) that are order-two. Thus P/Q is the othergroup of order four, and the lattices between Q and P are Zn = Qt

Ä(1, 0, . . . ) +Q

ä,

Qt(β+Q), and QtÄ(3

2 ,12 , . . . ,

12)+Q

ä. Thus when n is even there are five connected

complex Lie groups with Lie algebra so(2n,C). **

4. If G is an affine algebraic group, and g its Lie algebra, show that the canonical algebrahomormorphism U(g) → O(G)∗ identies U(g) with the set of linear functionals on O(G)whose kernel contains a power of the maximal ideal m = ker(eve).

5. Show that there is a unique Lie group over C with Lie algebra of type E8. Find the dimensionof its smallest matrix representation.

**We have shown already that there is a unique Lie group over C with Lie algebraE8, as the root- and weight-lattices of E8 agree, by problem 15 on Problem Set 5.I believe that its smallest nontrivial representation is the adjoint action E8 y e8;certainly this tensor-generates all representations, since the adjoint representa-tion tensor-generates the representations of the adjoint group. But I think thatto prove this the smallest representation requires material from the one day thissemester that I missed (Thanksgiving Eve). **

6. Construct a finite dimensional Lie algebra over C which is not the Lie algebra of any algebraicgroup over C. [Hint: the adjoint representation of an algebraic group on its Lie algebra isalgebraic.]

**Semisimple algebras and nilpotent algebras are always algebraically integrable,

202

the former by hard results, the latter almost trivially (Ado says any algebra is asubalgebra of gl(n), but on a nilpotent algebra exp : gl → GL is polynomial). Weshould look for a solvable but not nilpotent algebra. Luckily, there are a few ofthese even at dimension 3, c.f. Problem Set 4.

Let g be the three-dimensional algebra with basis X,Y, Z such that [X,Y ] = λY ,[X,Z] = Z, and [Y,Z] = 0, for some λ 6∈ Q. The center of g is trivial, so g ↪→ gl(3) as

X =

0 0 00 λ 00 0 1

, Y =

0 0 0−λ 0 00 0 0

, Z =

0 0 00 0 01 0 0

(PS6.18)

and so

g =

0 0 0b λa 0c 0 a

s.t. a, b, c ∈ C

(PS6.19)

and the algebraic closure of exp g is’exp g =

1 0 0r eλt 0s 0 et

s.t. r, s, t ∈ C

(PS6.20)

(exp g does not contain the matrices when s (resp. r) is non-zero but t (resp. λt)is a non-zero integer-multiple of 2πi, but the algebraic closure does contain thesematrices.) This is definitely a Lie subgroup of GL(3), but it is not algebraic ifλ 6∈ Q. In particular, it is not analytically closed as a subset of GL(3,C).

I have not shown that g does not have an algebraic integral — the irrational line inthe torus is not an algebraic group, but its Lie algebra also integrates to the one-dimensional Lie group(s). But let G be algebraic and connected with Lie(G) = g.Then G → GL(g) = GL(3,C) via the adjoint action, and the image is connected,hence generated by the exponential of the image of g, and so by construction theimage is ’exp g above. (We use the fact that for linear groups Ad(expx) = exp(adx).)However, for a linear algebraic group G, the adjoint action G → GL(Lie(g)) isalgebraic: if G ⊆ GL(n,C) ⊆ SL(n + 1,C), so Lie(g) ⊆ sl(n + 1), then G acts onLie(G) by matrix multiplication: Ad(X) · Y = XYX−1. But the map X 7→ X−1 ispolynomial on SL, as is matrix-multiplication, and so Ad : G → GL(Lie(G)) is arestriction of an algebraic map SL(n + 1) → GL(Lie(G)) to the subvariety G, andhence has algebraic image. **

203

Index

**, 5

abelian Lie algebra, 37adjoint action, 29Ado’s Theorem, 34, 49, 74, 76affine algebraic group, 118, 120affine type, 181affine variety, 117affine Weyl group, 182algebra, 39

associated graded, 40differential operators, 41filtered, 40Hopf, 118reductive, 95tensor, 36universal enveloping, 36

algebraically integrable, 122antipode, 52, 118associated graded algebra, 40associated graded module, 41associated graded vector space, 41atlas, 19augmented complex, 65

Baker-Campbell-Hausdorff formula, 42bialgebra, 40boosts, 160bracket, 9bracket polynomial, 149

Cartan matrix, 83, 102, 103indecomposable, 102

Cartan subalgebra, 88, 89Cartan’s Criteria, 61Cartan’s Theorem, 61Casimir operator, 64central series

lower, 53upper, 52

chain rule, 23

character, 114characteristic class, 66chart, 19charts

compatible, 19Chevellay group, 127classical groups, 17classical Lie algebras, 16, 78Clifford module, 126closed

Zariski, 88closed linear groups, 17coaction, 120coalgebra, 39coassociative, 118cohomology, 65compact real form, 17compatible charts, 19complete reducibility, 66complex, 65

augmented, 65complex conjugate, 14complex Lie group, 20comultiplication, 118concatenate, 47conjugate

complex, 14Hermitian, 14

connectedlocally path, 48locally simply, 48path, 47simply, 47

contractible, 48coordinates, 19

local, 19coroot lattice, 96coroots, 95covering space, 47coweight lattice, 96

204

Coxeter element, 189Coxeter groups, 106Coxeter number, 188

degenerate, 95derivation, 25, 143

point, 21derived algebra, 51derived series, 51, 52differential, 23differential operators, 41dominant integral weights, 112dominant morphism, 123dual root system, 100Dynkin diagram, 83, 102, 103

Engel’s Theorem, 54, 55Euclidean norm, 14evaluation, 118evaluation map, 117exponents, 188

family of vector fields, 31filtered algebra, 40filtered module, 41, 66filtered vector space, 41filtration, 37finite root system, 95finite-type, 110folding, 179free, 65free Lie algebra, 37free resolution, 65full rank, 123function

smooth, 20fundamental groupoid, 47fundamental theorem, 33fundamental weight, 114, 190

g-module, 37generator, 121graded Hopf algebra, 118group, 6

affine algebraic, 118classical, 17closed linear, 9, 17complex, 8, 20hyperoctahedral, 84Lie, 20linear algebraic, 8real, 20real C∞, 8real analytic, 8special linear, 14special orthogonal, 14special unitary, 14topological, 7unitary, 14

group action, 8group object, 7group scheme, 118, 127groupoid, 47

fundamental, 47

Hausdorff, 26height, 187, 188Hermetian conjugate, 14homology, 65homology indexing, 65homotopy, 47Hopf algebra, 8, 118

graded, 118Hopf ideal, 118hyperoctahedral group, 84

ideal, 51Hopf, 118

immersed submanifold, 22immersion, 22, 134indecomposable, 109integrable, 112integral curve, 26, 30irreducible, 66

Jordan decomposition, 91Jordan-Holder series, 55

205

Kac-Moody algbera, 107Killing form, 58

Langland’s Duality, 100left action, 28Levi, 72Levi subalgebra, 72Levi’s Theorem, 62Lie algebra, 25, 29

classical, 78Lie algebra homomorphism, 18Lie group, 20

complex, 20real, 20

Lie subalgebra, 11Lie subgroup, 44Lie’s Theorem, 56local coordinates, 19local isomorphism, 18locally path connected, 48locally simply connected, 48lower central series, 51, 53Lyndon word, 148

Malcev-Harish-Chandra theorem, 170manifold, 19, 20map

smooth, 20matrix

Cartan, 83matrix entries, 120matrix exponential, 9minimal roots, 81module

associated graded, 41Clifford, 126filtered, 41, 66

morphism, 117

necklaceprimitive, 148

nilpotent, 51, 52nilradical, 163norm

Euclidean, 14

octonions, 200open

Zariski, 88

path, 46path components, 47path connected, 47

locally, 48Poincare-Birkhoff-Witt, 37point derivation, 21, 119polynomial functions, 117positive matrix, 102positive root system, 100positive roots, 81primitive, 40, 118, 121primitive necklace, 148primitive word, 147product, 7, 20

quaternions, 14

radical, 56, 58rank, 81real form, 132real Lie group, 20reductive, 78, 126reductive algebra, 95regular, 88, 89representation, 36, 37, 64right action, 28right standard factorization, 149root, 80

minimal, 81positive, 81simple, 81, 96, 100

root datum, 126root lattice, 96root spaces, 80root system, 95–97

dual, 100isomorphism, 97positive, 100

206

rotations, 159

s/a/h, 21Schur’s Lemma, 68, 91semi-direct product, 51semisimple, 52series

derived, 52lower central, 53upper central, 52

Serre Relations, 109sheaf axioms, 20sheaf of functions, 20simple, 51simple root, 101simple roots, 81, 82, 96, 100simply connected, 47

locally, 48smooth, 118smooth map, 8, 20smooth vector field, 25solvable, 51, 52special linear group, 14, 15special orthogonal group, 14, 15special unitary group, 14string, 97strongly connected, 109subalgbra

Cartan, 88subalgebra

Cartan, 89Levi, 72

submanifoldimmersed, 22

substring, 185symmetrizable, 102symplectic group, 15

tangent, 20tangent space, 13tangent vector, 13, 21tensor algebra, 36triangular decomposition, 108

trivial extension, 71

unitary group, 14universal cover, 48universal enveloping algebra, 36upper central series, 52

varietyaffine, 117

vector field, 25family, 31smooth, 25

vector spaceassociated graded, 41filtered, 41

Verma module, 108, 112

weightfundamental, 114, 190

weight grading, 112weight lattice, 96weight space, 87Weyl, 68Weyl group, 83, 96, 97

affine, 182Whitehead, 68word

Lyndon, 148primitive, 147

Zariski closed, 88Zariski closure, 122Zariski open, 88Zassenhaus’ Extension Lemma, 75

207

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categorified.netcategorified.net/LieGroups.pdf · Lie Groups Prof. Mark Haiman notes by Theo Johnson-Freyd UC-Berkeley Mathematics Department Fall Semester 2008 Contents Contents